93. Binomials, Multinomials and Expansions Solutions

1. $$\left(x + \frac{1}{x}\right)^5$$

$$= {}^5C_0x^5 + {}^5C_1x^4\frac{1}{x} + {}^5C_2x^3\frac{1}{x^2} + {}^5C_3x^2\frac{1}{x^3} + {}^5C_4x\frac{1}{x^4} + {}^5C_5\frac{1}{x^5}$$

$$= x^5 + 5x^3 + 10x + \frac{10}{x} + \frac{5}{x^3} + \frac{1}{x^5}$$

2. $$(10.1)^5 = (10 + 0.1)^5$$

$$= 10^5 + {}^5C_110^4(0.1) + {}^5C_210^3(0.1)^2 + {}^5C_310^2(0.1)^3 + {}^5C_410(0.1)^4 + (0.1)^5$$

$$= 100000 + 5000 + 100 + 1 + 0.0005 + 0.00001$$

$$= 105101.000501$$

3. Let $$\sqrt{x - 1} = a$$

$$(x + \sqrt{x - 1})^6 = (x + a)^6$$

$$= x^6 + {}^6C_1x^5a + {}^6C_2x^4a^2 + {}^6C_3x^3a^3 + {}^6C_4x^2a^4 + {}^6C_5xa^5 + a^6$$

$$(x - \sqrt{x - 1})^6 = (x - a)^6$$

$$= x^6 - {}^6C_1x^5a + {}^6C_2x^4a^2 - {}^6C_3x^3a^3 + {}^6C_4x^2a^4 - {}^6C_5xa^5 + a^6$$

$$\therefore (x + \sqrt{x - 1})^6 + (x - \sqrt{x - 1})^6 = 2x^6 + 2{}^6C_2x^4a^2 + 2{}^6C_4x^2a^4 + 2a^6$$

$$= 2[x^6 + 15x^5 - 29x^3 + 12x^2 + 3x - 1]$$

4. Clearly $$(x + a)^n = A + B$$ and $$(x - a)^n = A - B$$

Thus, $$A^2 - B^2 = (x^2 - a^2)^n$$

5. Let us solve these two parts one by one

1. Given $$(7 + 4\sqrt{3})^n = \alpha + \beta$$

$$7 - 4\sqrt{3} = \frac{49 - 48}{7 + 4\sqrt{3}} = \frac{1}{7 + 4\sqrt{3}} < 1$$

$$\therefore (7 - 4\sqrt{3})^n = \beta_1 < 1$$

$$\alpha + \beta + \beta_1 = 2[7^n + {}^nC_27^{n - 2}48 + \ldots]$$

$$=~\text{an even integer}$$

$$\beta + \beta_1 =~\text{an even integer - a positive integer = an integer}$$

$$\because 0 < \beta < 1$$ and $$0 < \beta_1 < 1$$

$$\therefore 0 < \beta + \beta_1 < 2$$

$$\therefore \beta + \beta_1 = 1$$

$$\alpha + 1 =~\text{an even number}~\therefore \alpha =~\text{an odd number}$$

2. $$(\alpha + \beta)(1 - \beta) = (\alpha + \beta)\beta_1$$

$$= (7 + 4\sqrt{3})^n(7 - 4\sqrt{3})^n = 1$$

6. Let $$r^{th}$$ term contain $$\frac{1}{y^2}$$

$$t_r = {}^{10}C_{r - 1}y^{10 - r + 1}\left(\frac{c^3}{y^2}\right)^{r - 1}$$

$$={}^{10}C_{r - 1}y^{13 - 3r}c^{3r - 3}$$

$$\therefore 13 -3r - =2\Rightarrow r = 5$$

$$\therefore$$ Coefficient of $$y^-2 = {}^{210}c^{12}$$

7. $$(1 + 3x + 3x^2 + x^3)^{15} = [(1 + x)^3]^{15} = (1 + x)^{45}$$

$$\therefore$$ Coefficient of $$x^9 = {}^{45}C_9 = \frac{45!}{9!36!}$$

8. Let $$r^{th}$$ term be independent of $$x$$

$$t_r = {}^9C_{r -1}\left(\frac{3}{2}x^2\right){10 - r}\left(-\frac{1}{3x}\right)^{r - 1}$$

$$= (-1)^{r - 1}{}^9C_{r - 1}\left(\frac{3}{2}\right)^{10 - r}\frac{1}{3^{r - 1}}x^{21 - 3r}$$

$$\therefore 21 - 3r = 0 \Rightarrow r = 7$$

$$\therefore$$ Coefficient $$= \frac{7}{18}$$

9. $$(1 + x)^m\left(x + \frac{1}{x}\right)n = x^{-n}(1 + x)^{m + n}$$

Thus, the term which will have $$x^n$$ in $$(1 + x)^{m + n}$$ will be independent of $$x.$$

$$\therefore$$ Coefficient of $$x^n$$ in $$(1 + x)^{m + n} = {}^{m + n}C_n = \frac{(m + n)!}{m!n!}$$

10. Coefficient of $$x^{-1}$$ in $$(1 + 3x^2 + x^4)\left(1 + \frac{1}{x}\right)^8 =$$ coefficient of $$x^{-1}$$ in $$\left(1 + \frac{1}{x}\right)^8$$ + $$3 *$$ coefficient of $$x^{-3}$$ in $$\left(1 + \frac{1}{x}\right)^8$$ + coefficient of $$x^{-5}$$ in $$\left(1 + \frac{1}{x}\right)^8$$

$$r^{th}$$ term in the expansion of $$\left(1 + \frac{1}{x}\right)^8$$ is given by $$t_r = {}^8C_{r - 1}x^{1 - r}$$

When $$1 - r = 1, r = 2,$$ coefficient $$= {}^8C_1$$

When $$1 - r = -3, r = 4,$$ coefficient $$= {}^8C_3$$

When $$1 - r = -5, r = 6$$ coefficient $$= {}^8C_5$$

Required coefficient $$= {}^8C_1 + {}^8C_3 + {}^8C_5 = 222$$

11. $$a_{r - 1} = {}^{2n - 1}C_{r - 1}(-1)^{r - 1}$$

$$a_{2n - r} = {}^{2n - 1}C_{r - 1}(-1)^{r}$$

$$a_{r - 1} + a_{2n - r} = 0$$

12. $$t_r = {}^{10}C_{r - 1}x^{\frac{15 - 5r}{2}}k^{r - 1}$$

Since the term is independent of $$x, \therefore 15 - 5r = 0 \Rightarrow r = 3$$

$$t_3 = {}^{10}C_2k^2 = 405 \therefore k = \pm 3$$

13. $$t_k = {}^{n - 3}C_{k - 1}x^{n - 3k}$$

Since $$t_k$$ should contains $$x^{2r}, \therefore 2k = n - 3k$$

$$k = \frac{n - 2r}{3}$$

If $$n - 2r$$ is not divisible by $$3$$ then the expansion will have no term with $$x^{2r}$$

14. $$t_r = {}^nC_{r - 1}x^{an - (a + b)(r - 1)}$$

Since the term has to be independent of $$x, \Rightarrow an - (a + b)(r - 1) = 0$$

$$r = 1 + \frac{an}{a + b}$$

Thus, for $$r$$ to be an integer $$an$$ must be a multiple of $$a + b$$

15. $$\left(x + \frac{1}{x}\right)^7 = x^7 + {}^7C_1x^5 + {}^7C_2x^3 + {}^7C_3x + {}^7C_4x^{-1} + {}^7C_5x^{-3} + {}^7C_6x^{-5} + x^{-7}$$

$$= x^7 + 7x^5 + 21x^3 + 35x + 35x^{-1} + 21x^{-3} + 7x^{-5} + x^{-7}$$

16. This problem is simple and has been left as an exercise.

17. $$(1 + ax)^n = 1 + 8x + 24x^2 + \ldots = 1 + {}^nC_1ax + {}^nC_2a^2x^2 + \ldots$$

Comparing coefficients of $$x, an = 8$$

Comparing coefficients of $$x^2, \frac{n(n - 1)}{2}a^2 = 24$$

$$\Rightarrow \frac{64 - 8a}{2} = 24, \Rightarrow a = 2, n = 4$$

18. This problem is simple and similar to 3 and has been left as an exercise.

19. $$t_7 = {}^9C_6\left(\frac{4x}{5}\right)^3\left(-\frac{5}{2x}\right)^6$$

$$= \frac{10500}{x^3}$$

20. This problem is simple and similar to 3 and 18 and has been left as an exercise.

21. $$(0.99)^{15} = (1 - 0.01)^15$$

Since we have to evaluate only for $$4$$ decimal places considering first four terms will do.

First three terms $$= {}^{15}C_0 - {}^15C_1(.01) + {}^{15}C_2(.01)^2 + {}^{15}C_3(.01)^3 = 1 - .15 + .0105 - .000455 = 0.8600$$

22. $$(999)^3 = (1000 - 1)^3 = 1000^3 - 3. \frac{1000^2}{2} + 3.\frac{1000}{2} - 1$$

$$= 99700299$$

23. $$(0.99)^{10} = (1 - 0.01)^{10}$$

Since we have to evaluate only for $$4$$ decimal places considering first four terms will do.

First three terms $$= {}^{10}C_0 - {}^10C_1(.01) + {}^{10}C_2(.01)^2 - {}^{10}C_3(.01)^3 = 10 - .1 + 0.0045 - .000105 = .9044$$

24. This problem is simple and similar to 21 and 23 and has been left as an exercise.

25. $$A = {}^nC_0x^n + {}^nC_2x^{n - 2}a^2 + {}^nC_4x^{n - 4}a^4 + \ldots$$

$$B = {}^nC_1x^{n - 1}a + {}^nC_3x^{n - 3}a^3 + {}^nC_5x^{n - 5}a^5 + \ldots$$

$$(x + a)^{2n} - (x - a)^{2n} = 2[{}^{2n}C_1x^{2n - 1}a + {}^2nC_3x^{2n - 3}a^3]$$

$$4AB = 4{}^nC_0{}^nC_1x^{2n - 1}a + 4x^{2n - 3}a^3[{}^nC_0{}^nC_3 + {}^nC_1{}^nC_2 + \ldots]$$

Thus, we see that $$4AM = (x + a)^n - (x - a)^n$$

26. Let $$(5 + 2\sqrt{6})^n = \alpha + \beta$$ where $$\alpha$$ is a positive integer and $$beta$$ is a proper fracttion.

Also let, $$\gamma = (5 - 2\sqrt{6})^n$$

Now, $$5 - 2\sqrt{6} = \frac{5^2 - 4*6}{5 + 2\sqrt{6}} = \frac{1}{5 + 2\sqrt{6}} < 1$$

$$\therefore \gamma^n < 1$$

$$\alpha + \beta + \gamma = 2[5^n + {}^nC_25^{n - 2}6^2 + \ldots] =$$ An even number

$$\beta + \gamma =$$ An even number $$-$$ An interger = An integer

$$0 < \beta < 1$$ and $$0 < \gamma < 1$$

$$\therefore \beta + \gamma = 1$$

$$\therefore \alpha$$ is an off number.

27. This problem is simple and similar to 5. It has been left as an exercise.

28. $$t_r = {}^9C_{r - 1}(2x)^{9 - r + 1}\left(-\frac{3}{x}\right)^{r - 1}$$

$$= {}^9C_{r - 1}2^{9 -r + 1}(-3)^{r - 1}x^{9 - r + 1 - r + 1}$$

Now the power of $$x$$ should be $$1, \therefore 11 - 2r = 1 \Rightarrow r = 5$$

Coefficients of $$x$$ is $$= {}^9C_42^{5}(-3)^{4} = 2592{}^9C_4$$

29. $$t_r = {}^{11}C_{r - 1}(3x^2)^{11 - r + 1}(5x)^{1 - r}$$

$$= {}^{11}C_{r- 1}3^{12 - r}(5)^{r - 1}x^{24 - 2r + 1 - r}$$

$$7 = 25 - 3r \Rightarrow r = 6$$

$$\therefore$$ Coefficients of $$t_6 = {}^{11}C_53^65^5$$

30. $$t_r = {}^{20}C_{r - 1}(2x^2)^{20 - r + 1}(-x)^{1 - r}$$

$$= {-1}^{r - 1}{}^{20}C_{r - 1}2^{21 - r}x^{42 - 2r + 1 - r}$$

Since we need coefficients of $$x^9, \therefore 9 = 43 - 3r, \Rightarrow r = \frac{34}{3}$$

Since $$r$$ is not an integer, there is no term containing $$x^9$$ leading coefficient to be $$0.$$

31. $$t_r = {}^{15}C_{r - 1}(x^2)^{15 -r + 1}(3ax^{-1}){r - 1}$$

$$t-r = {}^{15}C_{r - 1}.(3a)^{r - 1}x^{32 - 2r + 1 - r}$$

Since we need coefficients of $$x^{24}, \therefore 24 = 33 - 3r$$

$$\Rightarrow r = 3$$

$$\therefore$$ Coefficient of $$x^{24} = 9a^2{}^{15}C_2$$

32. $$t_r = {}^9C_{r - 1}(x^2)^{9 - r + 1}(-3x^{-1})^{r - 1}$$

$$= {}^9C_{r - 1}(-3)^{r - 1}x^{20 - 2r + 1 - r}$$

Since we need coefficients of $$x^{9}, \therefore 9 = 21 - 3r, \Rightarrow r = 4$$

$$\therefore$$ Coefficent of $$x^9, = 27.{}^9C_3$$

33. $$t_r = {}^{11}C_{r - 1}(2x)^{11 -r + 1}\left(\frac{1}{3x^2}\right)^{r - 1}$$

$$= {}^{11}C_{r - 1}2^{12 - r}\frac{1}{3^{r - 1}}x^{12 - r + 2 - 2r}$$

Since we need coefficients of $$x^{-7}, \therefore -7 = 14 - 3r, \Rightarrow r = 7$$

$$\therefore$$ Coefficients of $$x^{-7}, = {}^{11}C_62^6\frac{1}{3^5}$$

34. $$r^{th}$$ term in the expansion of $$\left(ax^2 + \frac{1}{bx}\right)^{11}$$ is given by $$t_r = {}^{11}C_{r - 1}(ax^2)^{12 - r}\frac{1}{(bx)^{r - 1}}$$

$$t_r = {}^{11}C_{r - 1}a^{12 - r}b^{1 - r}x^{24 - 2r + 1 - r}$$

Since we need coefficient of $$x^7, 7 = 25 -3r, \Rightarrow r = 6$$

Coefficient of $$x^7, = {}^{11}C_5a^6b^{-5}$$

$$r^{th}$$ term in the expansion of $$\left(ax + \frac{1}{bx^2}\right)^{11}$$ is given by $$t_r = (-1)^{r - 1}.{}^{11}C_{r - 1}(ax)^{12 - r}\frac{1}{(bx^2)^{r - 1}}$$

$$t_r = (-1)^{r- 1}.{}^{11}C_{r - 1}a^{12 -r}b^{1 - r}x^{12 - r + 2 - 2r}$$

Since we need coefficient of $$x^{-7}, -7 = 14 -3r, \Rightarrow r = 7$$

Cofficient of $$x^{-7} = {}^{11}C_6a^5b^{-6}$$

Equating the coefficients we get $$ab = 1 \because {}^{11}C_5 = {}^{11}C_6$$

35. $$t_r = {}^{2n}C_{r - 1}x^{4n - 2r + 2}\frac{1}{x^{r - 1}}$$

$$t_r = {}^{2n}C_{r - 1}x^{4n - 3r + 3}$$

Since it is the term containing $$x^p, p = 4n - 3r + 3, \Rightarrow r = \frac{4n - p + 3}{3}$$

$$\therefore$$ Coefficients of $$x^p, = {}^{2n}C_{\frac{4n - p}{3}}$$

$$= \frac{2n!}{\left(\frac{4n -p}{3}\right)!\left(\frac{2n + p}{3}\right)}$$

36. Let us solve these one by one:

1. $$t_r = {}^{2n}C_{r - 1}x^{2n - r + 1}\frac{1}{x^{r - 1}}$$

Since the term has to be independent of $$x \Rightarrow 2n + 2 - 2r = 0 \Rightarrow r = n + 1$$

$$t_{n + 1} = {}^{2n}C_n = \frac{2n!}{n!n!}$$

2. $$t_r = {}^{15}C_{r - 1}(2x^2){16 - r}\frac{1}{x^{r - 1}}$$

$$\Rightarrow 33 - 3r = 0 \Rightarrow r = 11$$

$$t_{11} = {}^{15}C_10.2^5$$

3. $$t_r = {}^{10}C_{r - 1}\left(\sqrt{\frac{x}{3}}\right)^{10 - r + 1}\left(\frac{3}{2x^2}\right)^{r - 1}$$

$$\Rightarrow \frac{11 - r}{2} + 2 - 2r = 0 \Rightarrow r = 3$$

$$t_3 = {}^{10}C_2\frac{1}{3^4}\frac{3^2}{2^2} = {}^{10}C_2\frac{1}{6^2}$$

4. $$t_r = (-1)^{r - 1}.{}^{12}C_{r - 1}(2x^2)^{13 - r}\frac{1}{x^{r - 1}}$$

$$= (-1)^{r - 1}.{}^{12}C_{r - 1}2^{13 - r}x^{27 - 3r}$$

$$\Rightarrow 27 - 3r = 0 \Rightarrow r = 9$$

$$t_9 = {}^12C_82^4$$

5, 6, 7 and 8 are left as exercises.

37. $$t_r = {}^nC_{r - 1}x^{n - r + 1}\frac{1}{x^{2r - 21}}$$

$$= {}^nC_{r - 1}x^{n - 3r + 3}$$

For a term to be independent of $$x \Rightarrow r = \frac{n + 3}{3}$$

$$\therefore$$ Coefficient is $${}^nC_{\frac{n}{3}}$$

$$= \frac{n!}{\left(\frac{n}{3}\right)!\left(\frac{2n}{3}\right)!}$$

38. Coeff. of $$x^m$$ in $$(1 + x)^{m + n} = {}^{m + n}C_m$$

Coeff. of $$x^n$$ in $$(1 + x)^{m + n} = {}^{m + n}C_n$$

Clearly, both the coefficients are equal.

39. $$t_4 = {}^nC_3(px)^{n - 3}\frac{1}{x^3} = \frac{5}{2}$$

Since $$\frac{5}{2}$$ independent of $$x, n - 6 = 0 \Rightarrow n = 6$$

$${}^6C_3p^3 = \frac{5}{2} \Rightarrow p = \frac{1}{2}$$

40. Here $$n = 12,$$ which is even, therefore, $$\frac{12}{2} + 1$$ i.e. $$7{th}$$ term will be middle term.

$$t_7 = {}^{12}C_6x^6\left(-\frac{1}{2x}\right)^6 = \frac{231}{16}$$

41. Here $$n = 7,$$ which is odd, therefore, $$\frac{7 + 1}{2}$$ and $$\frac{7 + 3}{2}$$ i.e. $$4^{th}$$ and $$5^{th}$$ terms will be middle terms.

$$t_4 = {}^7C_3(2x^2){7 - 3}\left(-\frac{1}{x}\right)^3 = -560x^3$$

$$t_5 = {}^7C_4(2x^2)^{7 - 4}\left(-\frac{1}{x}\right)^2 = 280x^2$$

42. $$(1 - 2x + x^2)^n = (1 - x)^{2n}$$

Since $$2n$$ is even, therefore the middle term would be $$(n + 1)^{th}$$ term.

$$t_{n + 1} = {}^{2n}C_n1^{2n - n}(-x)^n = \frac{2n!}{n!n!}(-1)^nx^n$$

43. $$\because 2n$$ is even, the middle term would be $$(n + 1)^{th}$$ term.

$$t_{n + 1} = {}^{2n}C_nx^{2n - n}\frac{1}{x^n} = \frac{2n!}{n!n!}$$

$$= \frac{1.2.3.4\ldots 2n}{1.2.3\ldots n.(n!)}$$

$$= \frac{1.3.5\ ldots (2n - 1).2^{n}.1.2.3\ldots n}{1.2.3\ldots n.(n!)}$$

$$= \frac{1.3.5\ldots (2n - 1)}{n!}2^n$$

1. Clearly, middle term will have greatest coefficient which has been found in 44.

1. We have already found coefficient of middle term of $$(1 + x)^{2n}$$ which is $${}^{2n}C_n\frac{2n!}{n!n!}$$

Since $$2n - 1$$ is odd number we will have two middle terms for $$(1 + x)^{2n - 1},$$ which will be $$n^{th}$$ and $$(n + 1)^{th}$$ terms of the expansion.

Coefficient of $$t_n$$ in $$(1 + x)^{2n - 1} = {}^{2n - 1}C_{n - 1}$$

Coefficient of $$t_{n + 1} = {}^{2n - 1}C_n$$

Clearly, $${}^{2n - 1}C_n + {}^{2n - 1}C_{n - 1} = {}^{2n}C_n$$ by invoking properties of combinations.

2. Let us find these:

1. Since $$n = 20,$$ which is an even number, the middle term would be $$11^{th}$$ term.

$$t_{11} = {}^{20}C_{10}\left(\frac{2x}{3}\right)^{10} \left(-\frac{3}{2x}\right)^{10}$$

$$= {}^{20}C_{10}x^{10}y^{10}$$

2. Since $$n = 6,$$ an even number, the middle term would be $$4^{th}$$ term.

$$t_4 = {}^6C_3\left(\frac{2x}{3}\right)^3 \left(-\frac{3}{2x}\right)^3$$

$$= -20$$

3. Since $$n = 7,$$ an odd number, the middle terms would be $$4^{th}$$ and $$5^{th}$$ terms.

$$t_4 = {}^7C_3\frac{x^4}{y^4}.(-1)^3\frac{y^3}{x^3} = -35\frac{x}{y}$$

$$t_5 = {}^7C_4\frac{x^3}{y^3}.(-1)^4\frac{y^4}{x^3} = 25\frac{y}{x}$$

4. Since power of the expansion is $$2n,$$ which is an even number so $$(n + 1)^{th}$$ term would be middle term.

$$t_{n + 1} = {}^{2n}C_nx^n$$

5. $$(1 - 2x + x^2)^n = (1 - x)^{2n}$$ and like previous exercise $$t_{n + 1} = (-1)^n{}^{2n}C_nx^n$$ would be middle term.

3. Let $$(r + 1)^{th}$$ term be middle term then $$t_{r + 1} = {}^{2n + 1}C_r\frac{x^{2n - r + 1}}{y^{2n - r + 1}}\frac{y^r}{x^r}$$

$$t_{r + 1} = {}^{2n + 1}C_r\frac{x^{2n - 2r + 1}}{y^{2n - 2r + 1}}$$

Since $$2n + 1$$ is an odd number, therefore there will be two middle terms, $$(n + 1)^{th}$$ and $$(n + 2)^{th}$$

$$t_{n + 1} = {}^{2n + 1}C_n\frac{x}{y}$$

$$t_{n + 2} = {}^{2n + 1}C_n\frac{y}{x}$$

$$2n - 2r + 1,$$ which is power of $$x$$ and $$y$$ in general term, cannot be zero as both $$n$$ and $$r$$ are positive integers. Thus, no term is independent of $$x$$ and $$y.$$

4. Since exponent of expansion is $$2n$$ which is an even number, there will be one middle term and that term would be $$(n + 1)^{th}$$ term.

$$t_{n + 1} = {}^{2n}C_n(-1)^{n}\frac{x^n}{x^n}$$

$$= \frac{2n!}{n!n!} = \frac{2^n(1.2.3\ldots n)(1.3.5\ldots (2n - 1))}{n!n!}$$

$$= (-2^n)\frac{1.3.5\ldots (2n- 1)}{n!}$$

5. Coefficient of $$t_{2n + 1} = {}^{43}C_{2r}$$

Coefficient of $$t_{r + 2} = {}^{43}C_{r + 1}$$

Given $${}^{43}C_{2r} = {}^{43}C_{r + 1}$$

$$\therefore 2r + r + 1 = 43 \Rightarrow r = 14$$

6. Coefficient of $$r^{th}$$ term $$t_r = {}^{2n}C_{r - 1}$$

Coefficient of $$(r + 4)^{th}$$ term $$t_r = {}^{2n}C_{r + 3}$$

$$\Rightarrow 2n = r - 1 + r + 3 \Rightarrow r = n - 1$$

2nd Method Equal coefficient will be equidistant from mid term. In this case there is only one which is $$(n + 1)^{th}$$ term. So $$\frac{r + r + 4}{2} = n + 1 \Rightarrow r = n - 1$$

7. Following like previous exercise:

$$\Rightarrow 18 = 2r + 3 + r - 3 \Rightarrow r = 6$$

8. Following like previous exercise:

$$\Rightarrow 2r + 4 + r - 7 = 39 \Rightarrow r = 14$$

$${}^rC_{12} = \frac{14 * 13}{2} = 91$$

9. Following like previous exercise:

$$\Rightarrow 2n = 3r - 1 + r + 1 \Rightarrow r = \frac{n}{2}$$

10. Following like previous exercise:

$$\Rightarrow 2n = p + p + 2 \Rightarrow p = n - 1$$

11. Let $$r^{th}$$ and $$(r + 1)^{th}$$ have equal values for coefficients. Let $$C_r$$ represent cofficient for $$r^{th}$$ term and $$C_{r + 1}$$ for $$(r + 1)^{th}$$ term.

$$C_r = {}^{75}C_{r - 1}$$

$$C_{r + 1} = {}^{75}C_r3$$

Given $$C_r = C_{r + 1}$$

$$\frac{1}{75 - r + 1} = \frac{1}{r}$$

$$76 - r = r \Rightarrow r = 38$$

$$\therefore {}^{75}C_{37}$$ and $${}^{75}C_{38}$$ are equal coefficients.

12. Coefficient of $$(r + 1)^{th}$$ term in $$(1 + x)^{n + 1} = {}^{n + 1}C_r$$

Coefficient of $$r^{th}$$ term in $$(1 + x)^n = {}^nC_{r - 1}$$

Coefficient of $$(r + 1)^{th}$$ term in $$(1 + x)^n = {}^nC_r$$

Now we know that $${}^nC_{r - 1} + {}^nC_r = {}^{n + 1}C_{r + 1}$$

13. We have to find greatest term numerically.

$$therefore$$ greatest term in $$\left(7 -\frac{10}{3}\right)^{11} =$$ greatest term in $$\left(7 - \frac{10}{3}\right)^{11}$$

Let $$r^{th}$$ term be the greatest term in the expansion of $$\left(7 + \frac{10}{3}\right)^{11}$$

$$t_r = {}^{11}C_{r - 1}7^{11 -r + 1}\left(\frac{10}{3}\right)^{r - 1}$$

$$t_{r + 1} = {}^{11}C_r7^{11 - r}\left(\frac{10}{3}\right)^r$$

$$\frac{t_r}{t_{r + 1}} = \frac{21r}{(12 - r)}10$$

Also, $$\frac{t_{r - 1}}{t_r} = \frac{21(r - 1)}{13 - r}10$$

$$\because t_r$$ is the greatest term $$\therefore t_r \geq t_{r + 1}$$

$$\frac{21r}{(12 - r)10} \ geq 1 \Rightarrow r \geq \frac{120}{31}$$

Also, $$\frac{t_{r - 1}}{t_r} \leq 1$$

$$\Rightarrow r\leq \frac{151}{32}$$

Thus, $$r = 4$$

$$t_4 = \frac{440}{9}7^85^3$$

14. In any binomial expansion middle term has the greatest coefficient. Middle term in the expansion of $$(1 + 2x)^{2n}$$ is $$(n + 1)^{th}$$ term.

$$t_{n + 1} = {}^{2n}C_nx^n, t_{n + 1} = {}^{2n}C_{n + 1}x^{n + 1}, t_{n} = {}^{2n}C_{n - 1}x^{n - 1}$$

$$\frac{t_{n + 1}}{t_{n + 2}} = \frac{n + 1}{n}.\frac{1}{x} > 1 \Rightarrow x < \frac{n + 1}{n}$$

Similalrly $$\frac{t_{n + 1}}{t_n} = \frac{(n + 1)x}{n} > 1 \Rightarrow x > \frac{n}{n + 1}$$

Thus $$\frac{n}{n + 1} < x < \frac{n + 1}{n}$$

15. These are similar to problem 57 and have been left as an exercise.

16. This is similar to problem 58 and limit will be $$\frac{15}{16} < x < \frac{16}{15}$$.

17. Given expression $$= 6^{2n} - 35n - 1 = (6^2)^n - 35n - 1 = 36^n - 35n - 1$$

Clearly, $$35^2 = 1225$$ so we rewrite $$36$$ as $$1 + 35$$

$$(36)^n - 35n - 1 = (1 + 35)^n - 35n - 1$$

$$= 1 + 35n + {}^{35}C_2.35^2 + {}^{35}C_3.35^3 + \ldots + {}^{35}C_{35}35^n - 35n - 1$$

$$= {}^{35}C_2.35^2 + {}^{35}C_3.35^3 + \ldots + {}^{35}C_{35}35^n$$

All the terms contain powers euqal or greater than $$2$$ with base $$35$$ therefore whole expression is divisible by $$35^2$$ or $$1225$$

18. $$2^{4n} - 2^n(7n + 1) = (2^4)^n - 2^n.7n - 2^n$$

$$= 16^n - 2^n.7n - 2^n = (2 + 14)^n - 2^n.7n - 2^n$$

$$= {}^nC_02^n + {}^nC_12^{n - 1}14 + {}^nC_22^{n- 2}14^2 + {}^nC_32^{n - 3}14^3 + \ldots {}^nC_n14^n - 2^n.7n - 2^n$$

$$= 14^2[{}^nC_22^{n - 2} + {}^nC_32^{n - 3} + \ldots + {}^nC_n14^{n - 2}]$$

Thus, $$2^{4n} - 2^n(7n + 1)$$ is divisible by $$14^2$$

19. $$3^{4n + 1} + 16n - 3 = 3[3^{4n}] + 16n - 3 = 3[(1 + 80)^n - 1] + 16n$$

$$=3[{}^nC_0 + {}^nC_180 + {}^nC_280^2 + {}^nC_380^3 + \ldots + {}^nC_n80^n - 1] + 16n$$

$$= 256n + {}^nC_280^2 + {}^nC_380^3 + \ldots + {}^nC_n80^n$$

$$80^2$$ is divisible by $$256,$$ therefore above expression is divisible by $$256.$$

20. Let us solve these one by one:

1. $$4^n - 3n - 1 = (1 + 3)^n - 3n - 1 = 1 + 3n + {}^nC_23^2 + {}^nC_33^3 + \ldots + {}^nC_n3^n - 3n -1$$

$$= {}^nC_23^2 + {}^nC_33^3 + \ldots + {}^nC_n3^n$$

$$= 3^2[{}^nC_2 + {}^nC_33 + \ldots + {}^nC_n3^{n - 2}]$$

The above expression is divisible by $$9.$$

2. $$2^{5n} - 31n - 1 = (2^5)^n - 31n - 1 = 32^n - 31n - 1$$

$$= (1 + 31)^n - 31n - 1$$

$$= 1 + 31n + {}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n - 31 - 1$$

$$= 31^2[{}^nC_2 + {}^nC_331 + \ldots + {}^nC_n31^{n - 2}]$$

The above expression is divisible by $$961.$$

3. $$3^{2n + 2} - 8n - 9 = 3^2.3^{2n} - 8n - 9 = 9(1 + 8)^n - 8n - 9$$

$$= 9[1 + 8n + {}^nC_28^2 + {}^nC_38^3 + \ldots + {}^nC_n8^n] - 8n - 9$$

$$= 64n + 9[{}^nC_28^2 + {}^nC_38^3 + \ldots + {}^nC_n8^n]$$

The above expression is divisible by $$64$$

4. $$2^{5n + 5} - 31n - 32 = 2^5.(2^5)^n - 31n - 31 = 32(1 + 31)^n - 31n - 32$$

$$= 32[1 + 31n + {}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n] - 31n - 32$$

$$= 32[{}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n]$$

The above expression is divisible by $$31^2$$ i.e. $$961$$

5. $$3^{2n} - 1 + 24n - 32n^2 = (1 + 8)^n - 1 + 24n - 32n^2$$

$$= 1 + 8n + {}nC_28^2 + {}nC_38^3 + \ldots + {}nC_n8^n - 1 + 24n - 32n^2$$

$$= 32n + \frac{n(n - 1)}{2}8^2 + {}nC_38^3 + \ldots + {}nC_n8^n - 32n^2$$

$$= 32n + 64n^2 - 32n + {}nC_38^3 + \ldots + {}nC_n8^n$$

$$= 64n^2 + {}nC_38^3 + \ldots + {}nC_n8^n$$

Above expression is divisible by $$256$$ for $$n > 2$$

21. Let $$r^{th}, (r + 1)^{th}$$ and $$(r + 2)^{th}$$ terms are the three consecutive terms given in the problem.

$$t_r = {}^nC_{r - 1} = \frac{n!}{(r - 1)!(n - r + 1)!} = 165$$

$$t_{r + 1} = {}^nC_r = \frac{n!}{r!(n - r)!} = 330$$

$$t_{r + 2} \ {}^nC_{r + 1} = \frac{n!}{(r + 1)!(n - r - 1)!} = 462$$

$$\frac{t_r}{t_{r + 1}} = \frac{r}{n - r + 1} = \frac{1}{2} \Rightarrow 3r = n + 1$$

$$\frac{t_{r + 1}}{t_{r + 2}} = \frac{r + 1}{n - r} = \frac{330}{462}$$

$$12r = 5n - 7$$

Solving the two equations in $$n$$ and $$r,$$ we get, $$n = 11, r = 4.$$

22. Let $$a_1, a_2, a_3$$ and $$a_4$$ be coefficients of $$r^{th}, (r + 1)^{th}, (r + 2)^{th}$$ and $$(r + 3)^{th}$$ terms of the expansion $$(1 + x)^n$$

$$a_1 = {}^nC_{r - 1}, a_2 = {}^nC_r, a_3 = {}^nC_{r + 1}, a_4 = {}^nC_{r + 2}$$

$$\frac{a_2}{a_1} = \frac{n - r + 1}{r}$$

$$\frac{a_1}{a_1 + a_2} = \frac{r}{n + 1}$$

Substituting $$r + 1$$ instead of $$r$$ in previous equation, we get

$$\frac{a_2}{a_2 + a_3} = \frac{r + 1}{n + 1}$$

Similarly, $$\frac{a_3}{a_3 + a_4} = \frac{r + 2}{n + 1}$$

$$\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2(r + 1)}{n + 1} = \frac{2a_2}{a_2 + a_3}$$

23. $$\frac{t_2}{t_3} = \frac{{}^nC_1x^{n - 1}y}{{}^nC_2x^{n - 2}y^2} = \frac{240}{720}$$

$$\Rightarrow \frac{2x}{(n - 1)y} = \frac{1}{3}$$

$$\frac{t_3}{t_4} = \frac{{}^nC_2x^{n - 2}y^2}{{}^nC_3x^{n - 3}y^3} = \frac{720}{1080}$$

$$\Rightarrow \frac{3x}{(n - 2)y} = \frac{2}{3}$$

Dividing the two obtained equations

$$\frac{2x}{(n - 1)y}\frac{3x}{(n - 2)y} = \frac{1}{2}\frac{2}{3}$$

$$\frac{2(n - 2)}{3(n - 1)} = \frac{1}{2}$$

$$4n - 8 = 3n - 3 \Rightarrow n = 5$$

Putting this value in $$\frac{t_r}{t_{r + 1}},$$ we get

$$\frac{2}{5 -1}.\frac{x}{y} = \frac{1}{3}$$

$$y = \frac{3x}{2}$$

$$t_r = {}^nC_1x^{n - 1}y = \frac{3}{2}5x^5 = 240 \Rightarrow x = 2$$

$$\Rightarrow y = 3$$

24. Let $$n$$ be the index of the expansion and $$a, b, c$$ be the $$r^{th}, (r + 1)^{th}, (r + 2)^{th}$$ term respectively.

$$a = {}^nC_{r - 1}, b = {}^nC_r, c = {}^nC_{r + 1}$$

$$\frac{a}{b} = \frac{r}{n - r + 1}, \frac{b}{c} = \frac{r + 1}{n - r}$$

$$an + a = r(a + b), bn - c = r(b + c)$$

$$\Rightarrow (b + c)(an + a) = (a + b)(bn - c)$$

$$\Rightarrow n = \frac{2ac + b(a + c)}{b^2 - ac}$$

25. Coefficients are $$C_{14} = {}^nC_{13}, C_{15} = {}^nC_{14}, C_{16} = {}^nC_{15}$$

These are in A.P., so we can write

$$\frac{2.n!}{14!(n - 14)!} = \frac{n!}{13!(n - 13!)} + \frac{n!}{15!(n - 15)!}$$

Multiplying both sides by $$15!(n - 13)!,$$ we get

$$2.15(n - 13) = 15.14 + (n - 13).(n - 14)$$

$$\Rightarrow n^2 - 57n + 782 = 0$$

$$n = 23, 34$$

26. Let those three terms are $$r^{th}, (r + 1)^{th}$$ and $$(r + 2)^{th}$$ terms of the expansion.

$$\frac{t_{r- 1}}{t_r} = \frac{r}{n - r + 1} = \frac{56}{70} = \frac{4}{5}$$

$$9r - 4n = 4$$

$$\frac{t_{r}}{t_{r + 1}} = \frac{r + 1}{n - r} = \frac{70}{56}$$

$$9r - 5n = -4$$

$$\Rightarrow n = 8, r = 4$$

71, 72 and 73 are similar problems like ones we have solved and has been left as exercises.

1. Let the binomial expansion be $$(x + y)^n.$$

$$a = {}^nC_5x^{n - 5}y^5, b = {}^nC_6x^{n - 6}y^6, c = {}^nC_7x^{n - 7}y^7, d = {}^nC_8x^{n - 8}y^8$$

$$b^2 - ac = \left({}^nC_6x^{n - 6}y^6\right)^2 - {}^nC_5x^{n - 5}y^5{}^nC_7x^{n - 7}y^7$$

$$= \left(\frac{n!}{6!(n - 6)!}x^{n - 6}y^6\right)^2 - \frac{n!}{5!(n - 5)!}x^{n - 5}y^5 \frac{n!}{7!(n - 6)!}x^{n - 7}y^7$$

Similary $$c^2 - bd$$ can be computed and it can be shown that $$\frac{b^2 - ac}{c^2 - bd} = \frac{4a}{3c}$$

2. Let us solve these one by one.

1. We have to prove that $$\frac{a + b}{a}, \frac{b + c}{b}, \frac{c + d}{c}$$ are in H.P.

i.e. $$\frac{a}{a + b}, \frac{b}{b + c}, \frac{c}{c + d}$$ are in A.P.

Let $$a = {}^nC_r, b = {}^nC_{r + 1}, c = {}^nC_{r + 2}, d = {}^nC_{r + 3}$$

$$\frac{a}{a + b} = \frac{n!}{r!(n - r!)}\left(\frac{1}{\frac{n!}{r!(n - r!)}} + \frac{n!}{(r + 1)!(n - r - 1)!}\right)$$

$$= \frac{r + 1}{n + 1}$$

Similarly $$\frac{b}{b + c} = \frac{r + 2}{n + 1}$$

and, $$\frac{c}{c + d} = \frac{r + 3}{n + 1}$$

Clearly, these are in A.P.

2. Since $$\frac{a}{a + b}, \frac{b}{b + c}, \frac{c}{c + d}$$ are in A.P. from the previous part.

$$\frac{2b}{b + c} = \frac{a}{a + b} + \frac{c}{c + d}$$

Solving this further leads to the equality $$(bc + ad)(b - c) = 2(ac^2 - b^2d)$$

3. Since the coeficients of given terms are in A.P., we can write that

$$\frac{n!}{n - 4!} - \frac{2.n!}{5!(n - 5)!} + \frac{n!}{6!(n - 6!)} = 0$$

$$\frac{n!}{4!(n - 6!)}\left[\frac{1}{(n - 5)(n - 6)} -\frac{2}{5(n - 5) + \frac{1}{6.5}}\right]$$

Clearly, $$\frac{n!}{(n - 6!)}\neq 0$$

$$\Rightarrow n^2 - 23n + 132 = 0\Rightarrow n = 11, 12$$

4. Since the coeficients of given terms are in A.P., we can write that

$$\frac{2n!}{(2n - 1)!} + \frac{2.2n!}{2!(2n - 2)!} + \frac{2n!}{3!(2n - 3)!} = 0$$

$$2n - 2n(2n - 1) + \frac{n(2n - 1)(2n - 2)}{3} = 0$$

$$4n^3 - 18^2 + 14n = 0$$

$$\because n \neq 0, \Rightarrow 4n^2 - 9n + 7 = 0$$

5. Since the coeficients of given terms are in A.P., we can write that

$${}^nC_{r - 1} - 2{}^nC_r + {}^nC_{r + 1} = 0$$

$$\frac{n!}{(r - 1)!(n - r + 1)!} - 2\frac{n!}{r!(n - r)!} + \frac{n!}{(r + 1)!(n - r - 1)!} = 0$$

$$\frac{n!}{(r - 1)!(n - r - 1)!}\left[\frac{1}{(n - r)(n - r + 1)} - 2\frac{1}{r(n - r)} + \frac{1}{r(r + 1)}\right] = 0$$

Solving this gives the desired equation.

6. This problem is easy and similar to other problems and has been left as exercise.

7. Given series is $$C_1 + 2.C_2 + 3.C_3 + \ldots + n.C_n$$

Its $$r^{th}$$ term $$t_r = r.{}^nC_r = n.{}^{n - 1}C_{r -1}$$

$$C_1 + 2.C_2 + 3.C_3 + \ldots + n.C_n = \sum_{r = 1}^nr.{}^nC_r$$

$$= n\sum_{r=1}^n{}^{n - 1}C_{r - 1} = n.2^{n - 1}$$

Calculus Method: This method requires knowledge of calculus.

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Differentiating both sides w.r.t. $$x$$

$$n.(1 + x)^{n - 1} = {}^nC_1 + 2{}^nC_2x^2 + \ldots + n{}^nCnx^{n - 1}$$

Putting $$x = 1,$$ we get

$$n.2^{n - 1} = {}^nC_1 + 2{}^nC_2 + \ldots + n.{}^nC_n$$

8. Given series is $$C_0 + 2.C_1 + 3.C_2 + \ldots + (n + 1).C_n$$

Its $$r^{th}$$ term $$t_r = r{}^nC_{r - 1}$$

$$= (r - 1 + 1){}^nC_{r - 1} = (r - 1){}^nC_{r - 1} + {}^nC_{r - 1}$$

$$= n{}^{n - 1}C_{r - 2} + {}^nC_{r - 1}[\because (r - 1){}^nC_{r - 1} = n{}^{n - 1}C_{r - 2}]$$

Thus, $$\sum_{r = 1}^{n + 1}t_r = n.2^{n - 1} + 2^n = (n + 2)2^{n - 1}$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Multiplying both sides by $$x,$$ we get

$$x(1 + x)^n = {}^nC_0x + {}^nC_1x^2 + {}^nC_2x^3 + \ldots + {}^nC_nx^{n + 1}$$

Differentiating w.r.t. $$x,$$ we get

$$[(1 + x)^n + nx(1 + x)^{n - 1}] = {}^nC_0 + 2{}^nC_1x + 3{}^nC_2x^2 + \ldots + (n + 1){}^nC_nx^n$$

Substituting $$x = 1,$$ we get

$$C_0 + 2.C_1 + 3.C_2 + \ldots + (n + 1).C_n = (n + 2).2^{n - 1}$$

9. $$t_r = (2r - 1){}^nC_{r - 1} = 2(r - 1){}^nC_{r - 1} + {}^nC_{r - 1}$$

$$=2n {}^{n - 1}C_{r - 2} + {}^nC_{r - 1}$$

$$\sum_{r = 1}^{n + 1}t_r = \sum_{r = 1}^{n + 1}2n {}^{n - 1}C_{r - 2} + \sum_{r = 1}^{n + 1}{}^nC_{r - 1}$$

$$= 2n({}^{n - 1}C_0 + {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + {}^{n - 1}C_{n - 1}) + ({}^nC_0 + {}^nC_1 + {}^nC_2 + \ldots + {}^nC_n)$$

$$= 2n.2^{n - 1} + 2^n = 2^n(n + 1)$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Substituting $$x = x^2$$ and multiplying both sides with $$x$$

$$x(1 + x^2)^n = C_0x + C_1x^3 + C_2x^5 + \ldots + C_nx^{2n + 1}$$

Differentiating both sides w.r.t $$x$$ and substituting $$x = 1,$$ we get the desired equation as earlier.

10. Given series is $$C_1 - 2.C_2 + 3.C_3 - 4.C_4 + \ldots + (-1)^nn.C_n = 0$$

$$t_r = (-1)^{r - 1}r.{}^nC_r = (-1)^{r - 1}n.{}^{n - 1}C_{r - 1}$$

$$\sum_{r = 1}^{n}t_r = \sum_{r = 1}^{n}(-1)^{r - 1}n.{}^{n - 1}C_{r - 1}$$

$$n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n - 1}){}^{n - 1}C_{r - 1} = n(1 - 1)^{n - 1} = 0$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Differentiating w.r.t. $$x$$ and substituting $$x = -1$$

$$n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n - 1}){}^{n - 1}C_{r - 1} = n(1 - 1)^{n - 1} = 0$$

11. Given series is $$C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}$$

$$t_r = \frac{{}^nC_{r - 1}}{r}$$

$$C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \sum_{r = 1}^n \frac{{}^nC_{r - 1}}{r} = \sum_{r = 1}^n\frac{{}^{n + 1}C_r}{n + 1}\left[\because \frac{{}^nC_{r - 1}}{r} = \frac{{}^{n + 1}C_r}{n + 1}\right]$$

$$= \frac{1}{n + 1}({}^{n + 1}C_1 + {}^{n + 1}C_2 + {}^{n + 1}C_{n + 1})$$

$$=\frac{2^{n + 1} - 1}{n + 1}$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Integrating w.r.t $$x$$ between limits $$0$$ and $$1,$$ we get

$$\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^1 = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^1$$

$$\Rightarrow C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}$$

12. Given series is $$C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n\frac{C_n}{n + 1}$$

$$t_r = (-1)^{r - 1}\frac{{}^nC_{r - 1}}{r}$$

$$\therefore C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n\frac{C_n}{n + 1} = \sum_{r = 1}^{n + 1}(-1)^{r - 1}\frac{{}^nC_{r - 1}}{r} = \sum_{r = 1}^{n + 1}(-1)^{r - 1}\frac{{}^{n + 1}C_r}{n + 1}\left[\because \frac{{}^nC_{r - 1}}{r} = \frac{{}^{n + 1}C_r}{n + 1}\right]$$

$$= \frac{1}{n + 1}[{}^{n + 1}C_1 - {}^{n + 1}C_2 + {}^{n + 1}C_3 - \ldots + (-1)^n.{}^{n + 1}C_{n + 1}]$$

$$= \frac{1}{n + 1}[-({}^{n + 1}C_0 + {}^{n + 1}C_1 - {}^{n + 1}C_2 + {}^{n + 1}C_3 - \ldots + (-1)^{n+1}.{}^{n + 1}C_{n + 1}) + {}^{n + 1}C_0]$$

$$= \frac{1}{n + 1}\left[-(1 - 1)^{n + 1} + {}^{n + 1}C_0\right] = \frac{1}{n + 1}$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Integrating w.r.t $$x$$ between limits $$0$$ and $$1,$$ we get

$$\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^{-1} = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^{-1}$$

$$0 - \frac{1}{n + 1} = -C_0 + \frac{C_1}{2} - \frac{C_3}{3} + \ldots + (-1)^{n + 1}\frac{C_n}{n + 1}$$

$$\Rightarrow C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n \frac{C_n}{n + 1} = \frac{1}{n + 1}$$

13. Given series is $$\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots$$

$$t_r = \frac{{}^nC_{2r - 1}}{23} = \frac{{}^nC_{2r - 1}}{(2r - 1) + 1} = \frac{{}^{n + 1}C_{2r}}{n + 1}$$

$$\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots = \sum_{r = 1} = \frac{1}{n + 1}\sum_{r = 1}{n + 1}C_{2r} = \frac{2^n - 1}{n + 1}$$

Calculus Method:

Adding the results of 84 and 85, we get

$$2\left[\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots\right] = \frac{2^{n + 1} - 1 - 1}{n + 1} = \frac{2(2^n - 1)}{n + 1}$$

$$\Rightarrow \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots = \frac{2^n - 1}{n + 1}$$

14. Given series is $$2.C_0 + 2^2\frac{C_1}{2} + 2^3\frac{C_2}{3} + \ldots + 2^{n + 1}\frac{C_n}{n + 1}$$

$$t_r = 2^r\frac{{}^nC_{r - 1}}{r} = 2^r\frac{{}^{n + 1}C_r}{n + 1}$$

Now $$2.C_0 + 2^2\frac{C_1}{2} + 2^3\frac{C_2}{3} + \ldots + 2^{n + 1}\frac{C_n}{n + 1} = \sum_{r = 1}^{n + 1} 2^r\frac{{}6{n + 1}C_r}{n + 1}$$

$$= \frac{1}{n + 1}[\{{}^{n + 1}C_0 + 2.{}^{n + 1}C_1 + 2^2.{}^{n + 1}C_2 + \ldots + 2^{n + 1}{}^{n + 1}C_{n + 1}\} - {}^{n + 1}C_0]$$

$$= \frac{1}{n + 1}[(1 + 2)^{n + 1} - 1] = \frac{3^{n + 1} - 1}{n + 1}$$

Calculus Method:

$$(1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n$$

Integrating between limits $$0$$ and $$2,$$ we get the desired result.

15. $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

$$(x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n$$

Multiplying these two, we get

$$(1 + x)^{2n} = (C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)$$

Coefficient of $$x^{n + r}$$ on R.H.S $$= C_0C_r + C_1C_{r + 1} + \ldots + C_{n - r}C_n$$

Coefficient of $$x^{n + r}$$ on L.H.S. $$= {}^{2n}C_{n + r} = \frac{(2n)!}{(n + r)!(n - r)!}$$

Equating coefficients we get desired result.

16. From 88 recall that $$(1 + x)^{2n} = (C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)$$

Equating coefficients of $$x^n,$$ we get

$$C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 = \frac{(2n)!}{n!n!}$$

17. $$t_r = r.\frac{{}^nC_r}{{}^nC_{r - 1}} = n - r + 1$$

$$\sum_{r=1}^n = \frac{n(n + 1)}{2}$$

18. $$\{(1 + x)^n\}^2 = (1 + x)^{2n}$$

$$({}^nC_0x + {}^nC_1x + {}^nC_2 + \ldots + {}^nC_n)^2 = ({}^{2n}C_0 + {}^{2n}C_1x + {}^{2n}C_2x + \ldots + {}^{2n}C_{2n}x^{2n})$$

Substituting $$x = 1,$$ we get desired equation.

19. This problem is same as previous one, just that instead of $$2, 5$$ has been used.

20. $$t_r = (4r + 1){}^nC_{r - 1} = 4n{}^{n - 1}C_{r - 1} + {}^nC_{r - 1}$$

$$C_0 + 5.C_1 + 9.C_2 + \ldots + (4n + 1).C_n = \sum_{r = 0}^nt_r$$

$$=4n \sum_{r = 0}^n{}^{n - 1}C_{r - 1} + \sum_{r = 0}^n{}^nC_{r - 1}$$

$$= 4n.2^{n - 1} + 2^n = (2n + 1)2^n$$

21. L.H.S. $$1 - (1 + x)C_1 + (1 + 2x)C_2 - (1 + 3x)C_3 + \ldots$$

$$= 1 - C_1 + C_2 - C_3 + \ldots -x(C_1 - 2C_2 + 3C_3 + \ldots)$$

We know that $$(1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + \ldots$$

Substituting $$x = -1,$$ we get

$$0 = = 1 - C_1 + C_2 - C_3 + \ldots$$

Thus our original expression becomes $$-x(C_1 - 2C_2 + 3C_3 + \ldots) = 0$$

$$C_1 - 2C_2 + 3C_3 + \ldots = 0$$

$$t_r = (-1)^{r - 1}r.C_r$$

$$C_1 - 2C_2 + 3C_3 + \ldots = \sum_{r = 1}(-1)^{r - 1}r.C_r = n\sum_{r = 1}{}^{n- 1}C_{r - 1}$$

$$= n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n -1}{}^{n - 1}C_{n - 1})$$

$$= n(1 - 1)^{n - 1} = 0$$

22. $$t_r = (4r - 1)C_{r - 1} = 4rC_{r - 1} - C_{r - 1} = 4n{}^{n - 1}C_{r - 1} - C_{r - 1}$$

$$3.C_1 + 7.C_2 + 11.C_3 + \ldots + (4n - 1)C_n = 4n \sum_{r = 1}^{n}{}^{n - 1}C_{r - 1} - \sum_{r = 1}^n{}^nC_{r - 1}$$

$$= (2n - 1)2^n + 1$$

Problem no. 96 and 97 are similar to what we have solved and have been left as exercise.

1. Let $$(1 + x - 3x^2)^{2163} = a_0 + a_1x + a_2x^2 + \ldots +a_{6489}^{6489}$$

Substituting $$x = 1,$$ we get

$$a_0 + a_1 + a_2 + \ldots + a_{6489} = (-1)^2163 = -1$$

2. This problem can be solved by substituting $$1, \omega, \omega^2$$ for $$x$$ and adding. it has been left as an exercise.

3. $$t_{r + 1} = {}^{10}C_r2^{\frac{10 - r}{2}}3^{\frac{r}{5}},$$ where $$r = 0, 1, 2, \ldots, 10$$

For rational terms $$r =$$ a multiple of $$5 = 0, 5, 10$$

$$10 - r =$$ a multiple of $$2 = 0, 2, 4, 6, 8, 10$$

For both the only common values are $$r = 0, 10$$

$$\therefore$$ sum of rational terms $$= t_1 + t_{11}$$

$$= 41$$

4. $$\frac{2^{4n}}{15} = \frac{16^n}{15} = \frac{(1 + 15)^n}{15}$$

$$= \frac{1 + {}^nC_1.15 + {}^nC_2.15^2 + \ldots + {}^nC_n.15^n}{15}$$

Except first term all others are mutliple of $$15$$ so clearly fractional part $$= \frac{1}{15}$$

5. Let $$(\sqrt{3} + 1)^{2n} = p + f,$$ where $$p$$ is the integral part and $$f$$ is the fractional part i.e. $$0<f<1$$

Integer just above $$(\sqrt{3} + 1)^{2n} = p + 1$$

Now $$(\sqrt{3} + 1)^{2n} = [(\sqrt{3} + 1)^2]^n = 2^n(2 + \sqrt{3})^n$$

$$p + f = 2^n(2 + \sqrt{3})^n$$

Also, $$0 < \sqrt{3} - 1 < 1 \Rightarrow 0 < (\sqrt{3} - 1)^{2n} < 1$$

Let $$f_1 = \sqrt{3} - 1)^{2n} = 2^n(2 - \sqrt{3})^n, \Rightarrow 0 < f_1 < 0$$

$$p + f + f_1 = 2^n.2[2^n + {}^nC_22^{n - 2}.3 + \dots]$$

$$= 2^{n + 1}.$$ an integer = an even integer

$$f + f_1 =$$ Even integer - p = an odd integer

Also, $$0 < f + f_1 < 2$$

Clearly, $$f + f_1 = 1$$

$$\Rightarrow p + 1 = 2^{n + 1}.$$ an integer

Thus, integer next to $$(\sqrt{3} + 1)^{2n}$$ is divisible by $$2^{n + 1}$$

6. Let $$p$$ be the integral part of $$R,$$ then $$[R] = p$$

Since $$f = R - [R] = R - p \therefore 0 < f < 1$$

and $$R = p + f$$

$$p + f = (5\sqrt{5} + 11)^{2n + 1}$$

Let $$f_1 = (5\sqrt{5} - 11)^{2n + 1}$$

We observe that $$5\sqrt{5} - 11 = \frac{4}{5\sqrt{5} + 11}$$

$$\therefore 0 < f_1 < 1$$

$$p + f - f_1 = 2[{2n + 1}C_1.(5\sqrt{5})^{2n}.11 + {2n + 1}C_3.(5\sqrt{5})^{2n - 2}.11^3 + \ldots]$$

$$=$$ an even number

$$f - f_1 =$$ en even number $$- p =$$ an integer

$$-1 < f - f_1 < 1$$

Thus, we can say that $$f - f_1 = 0 \Rightarrow f = f_1$$

$$\therefore Rf = Rf_1 = r^{2n + 1}$$

7. Let $$x = 101^{50}$$ and $$y = 100^{50} + 99^{50}$$

$$101^{50} - 99^{50} = 100^{50} + 2[{}^{50}C_3.100^{47} + \ldots + {}^{50}C_{49}100]$$

$$= 100^{50} +$$ a positive number

$$101^{50} - 99^{50} > 100^{50}$$

$$101^{50} > 100^{50} + 99^{50}$$

8. $$t_1 = \sum_{r = 0}^n (-1)^r{}^nC_r\left(\frac{1}{2}\right) = \left(1 - \frac{1}{2}\right)^n = \frac{1}{2^n}$$

Similarly, $$t_2 = \frac{1}{2^{2n}}$$

and $$t_r = \frac{1}{2^{3n}}$$

$$\therefore$$ required sum $$= \frac{1 - \frac{1}{2^{mn}}}{2^n - 1}$$

9. $$32^{32} = (2 + 30)^{32} = 2^{32} + 30k,$$ where $$k \in N$$

Therefore last digits in $$32^{32} =$$ last digits in $$2^{32}$$

$$2^{32} = (2^5)^6.2^2 = 32^6.4 = (2^6 + 30r)4, r\in N$$

Last digit in $$2^6.4 =$$ last digit in $$64.4 = 6$$

$$\therefore$$ last digit in $$32^{32} = 6$$

10. Let $$n = 2m,$$ where $$m$$ is a positive integer, then $$k = 3m$$

L.H.S. $$= \sum_{r = 1}^{3m}(-3)^{r - 1}.{}^{6m}C_{2r - 1}$$

$$={}^{6m}C_1 - {}^{6m}C_3.3 + {}^{6m}C_5.3^2 + \ldots + (-1)^{3m - 1}{}^{6m}C_{6m - 1}.3^{3m - 1}$$

$$= \frac{1}{\sqrt{3}}[{}^{6m}C_1.\sqrt{3} - {}^{6m}C_3.(\sqrt{3})^3 + {}^{6m}C_5(\sqrt{5})^5 - \ldots + (-1)^{3m - 1}.{}^{6m}C_{6m - 1}(\sqrt{3})^{6m - 1}]$$

Also, $$(-1)^{3m - 1} = -i(i)^{6m - 1}$$

$$(1 + \sqrt{3}i)^{6m} = [1 - {}^6mC_2(\sqrt{3})^2 + {}^{6m}C4(\sqrt{3})^4 + \ldots] + i[{}^{6m}C_1.\sqrt{3} - {}^{6m}C_3(\sqrt{3})^3 + \ldots + {}^{6m}C_{6m -1}(i)^{6m - 2}(\sqrt{3})^{6m - 1}]$$

However, $$(1 + \sqrt{3}i)^6m = \left[2\left(\cos \frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\right]^{6m} = 2^{6m}[\cos 2m\pi + \sin 2m\pi]$$

$$= 2^{6m}(1 + 0) = 2^{6m}$$

Thus, coefficient of imaginary part is $$0,$$ which proves the desired result.

11. $$t_0 = a^n, t_1 = {}^nC_1a^{n - 1}x, t_2 = {}^nC_2a^{n - 2}x^2, ...$$

Substituting $$x = ix,$$ we get

$$(a + ix)^n = a^n + {}^nC1a^{n - 1}.(ix) + {}nC_2a^{n - 2}(ix)^2 + \ldots + {}^nC_n(ix)^n$$

$$=(t_0 - t_2 + t_4 - \ldots) + i(t_1 - t_3 + t_5 - \ldots)$$

Taking modulus and then squaring, we get

$$(t_0 - t_2 + t_4 - \ldots)^2 + (t_1 - t_3 + t_5 - \ldots)^2 = (a^2 + x^2)^n$$

12. Putting $$x = 1,$$ we get

$$3^n = a_0 + a_1 + a_2 + a_3 + \ldots + a_{2n}$$

13. Putting $$x = -1,$$ we get

$$1^n = a_0 - a_1 + a_2 - a_3 + \ldots +(-1)^{2n}a_2n$$

$$1^n = a_0 - a_1 + a_2 - a_3 + \ldots +a_2n$$

14. Substituting $$x = 1, \omega, \omega^2$$ and adding them we get the desired result.

15. $$S_n = \frac{1 - q^{n + 1}}{1 - q}$$

$$S_n^{'} = \frac{2^{n + 1} - (q + 1)^{n + 1}}{(1 - q).2^n}$$

Now, $${}^{n + 1}C_1 + {}^{n + 1}C_2.S_1 + {}^{n + 1}C_3.S_2 + \ldots + {}^{n + 1}C_{n + 1}.S_n$$

$$= {}^{n + 1}C_1\frac{1 - q}{1 - q} + {}^{n + 1}C_2\left(\frac{1 - q^2}{1 - q}\right) + {}^{n + 1}C_3\left(\frac{1 - q^3}{1 - q}\right) + \ldots + {}^{n + 1}C_{n + 1}\left(\frac{1 - q^{n + 1}}{1 - q}\right)$$

$$= \frac{1}{1 - q}[({}^{n + 1}C_1 + {}^{n + 1}C_2 + \ldots + {}^{n + 1}C_{n + 1}) -({}^{n + 1}C_1.q + {}^{n + 1}C_2.q^2 + \ldots + {}^{n + 1}C_{n + 1}.q^{n + 1})]$$

$$= \frac{1}{1 - q}[2^{n + 1}- (1 + q)^{n + 1}] = 2^nS_n^{'}$$

16. $$(\sqrt[4]{9} + \sqrt[6]{8})^{1000} = (\sqrt{3} + \sqrt{2})^{1000}$$

Clearly, terms $$0, 2, 4, \ldots, 1000$$ will be rational. Thus, total no. of rational terms is $$501.$$

17. Clearly, terms which are divisible by L.C.M. of $$3$$ and $$5$$ i.e. $$15$$ will be rational numbers. Such terms are first and last term.

Sum of rational terms $$= {}^{15}C_02^5 + {}^{15}C_03^3 = 59$$

18. $$t_3 = {}^5C_2x^4(\log_{10}x)^2 = 1000,000$$

Clearly $$x = 10$$

19. $$x^3 + 3x^2 - 5 + \frac{3}{x^2} - \frac{1}{x^3}$$

20. Since coefficients of second, third and fourth terms are in A.P., we can write

$$2{}^mC_2 = {}^mC_1 + {}^mC_3$$

$$m(m - 1) = m + \frac{m(m - 1)(m - 2)}{6}$$

$$\Rightarrow m^2 - 9m + 14 = 0 \Rightarrow m = 2, 7$$ but since we are dealing with sixth term $$m$$ will be 7 and we will need to discard $$2.$$

$$t_6 = {}7C_5(2^{\log(10 - 3^x)}).2^{(x - 2)\log 3} = 21$$

$$\Rightarrow 2^{\log(10 - 3^x) + (x - 2)\log 3} = 1$$

$$\log [10 - 3^x + (x - 2)\log 3] = 2$$

$$10 - 3^x + (x - 2)\log 3 = 100$$

Solving this yields

$$\Rightarrow x = 0, 2$$

21. This problem is similar to previous one and has been left as an exercise.

22. Given $$\frac{1}{(81)^n} - \frac{10}{(81)^n}{}^{2n}C_1 + \frac{10^2}{(81)^n}{}^{2n}C_2 - \frac{10^3}{(81)^n}{}^{2n}C_3 + \ldots + \frac{10^{2n}}{(81)^n} = 1$$

L.H.S. $$= \frac{1}{81^n}(1 - 10)^{2n}$$

$$= \frac{1}{81^n}(-9^2)^n = \frac{1}{81^n}81^n = 1$$

23. $$\because {}^nC_r = {}^nC){n - r}$$

$$\therefore S_n = {}^nC_0 - {}^nC_1\frac{2}{3} + {}^nC_2\left(\frac{2}{3}\right)^2 - \ldots + (-1)^n\left(\frac{2}{3}\right){}^nC_n$$

$$= \left(1 - \frac{2}{3}\right)^n = \frac{1}{3^n}$$

$$\therefore \lim_{n \to \infty}S_n = 0$$

24. This problem can be solved like 103 and has been left as an exercise.

25. $$\sum_{r = 0}^n(-1)^r.{}^nC_r\left[\frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{w^{3r}} + \ldots~\text{up to}~n~\text{terms}\right]$$

$$= \left(1 - \frac{1}{2}\right)^n + \left(1 - \frac{1}{4}\right)^n + \left(1 - \frac{1}{8}\right)^n + \ldots n$$ terms

$$= \frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \ldots n$$ terms

Summing this G.P. yields

$$= \frac{1}{2^n - 1} - \frac{1}{2^{n^2}(2^n - 1)}$$

26. $$17^{256} = 289^{128} = (290)^{128}$$

$$= 1000m + {}^{128}C_{126}290^2 - {}^{128}C_{127}290 + 1 [m \in I]$$

$$= 1000(m + 683527) + 681$$

Thus the digits are $$6, 8$$ and $$1.$$

27. $$(n + 1)^n = n^n\left(1 + \frac{1}{n}\right)^n$$

Upon expansion(as shown in following problem) you will find that $$\left(1 + \frac{1}{n}\right)^n < n$$

Thus, $$n^{n + 1} > (n + 1)^n$$

28. $$\left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \ldots$$

$$< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!}$$

$$< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n - 1}}$$

$$= 3 - \frac{1}{2^{n - 1}}$$

Thus, we have proven the desired inequality.

29. We will make use of the fact that $$x^n - y^n$$ is divisible by $$x - y$$

$$(1992^{1998} - 1955^{1998}) - (1938^{1998} - 1901^{1998})$$ is divisible by $$37$$

$$(1992^{1998} - 1938^{1998}) - (1955^{1998} - 1901^{1998})$$ is divisible by $$54$$

$$\therefore 1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}$$ is divisible by $$37*54$$ i.e. $$1998$$

30. Given expression is $$(50 + 3)^{53} - {30+ 3}^{33}$$

$$= 50p + {}^{53}C_{53}3^{53} - 30q - {}^{33}C_{33}3^{33}, p, q \in I$$

Thus, now we have to prove that $$3^{53} - 3^{33}$$ is divisible by $$10$$

$$3^{33}(3^{20} - 1)$$ now if you see carefully $$3^{20} = 81^{5}$$ which will awlays have last digit as 1. Thus, $$3^{20} - 1$$ will be always divisible by $$10$$ making out original expression also divisible by 10.

31. $$(1 + x)^{m + 1} = {}^{m + 1}C_0 + {}^{m + 1}C_1x + {}^{m + 1}C_2x^2 + \ldots + {}^{m + 1}C_mx^m + {}{m + 1}C_{m + 1}x^{m + 1}$$

$$(1 + x)^{m + 1} - 1 - x^{m + 1} = {}^{m + 1}C_1x + {}^{m + 1}C_2x^2 + \ldots + {}^{m + 1}C_mx^m$$

Substituting $$x = 1, 2, 3, 4, \ldots, n$$ in the above expression and adding, we get

$$= (n + 1)^{m + 1} - (n + 1)$$

32. $$\sum_{i = 1}^k\sum_{k = 1}^n{}^nC_k{}^kC_i$$

$$= \sum_{k = 1}^n({}^nC_k).({}^kC_1) + \sum_{k = 1}^n({}^nC_k).({}^kC_2) + \ldots + {}^nC_n.{}^nC_n$$

$$= {}^nC_1.{}^1C_1 + {}^nC_2({}^2C_1 + {}^2C_2) + \ldots + {}^nC_n({}^nC_1 + {}^nC_2 + \ldots + {}^nc_n)$$

$$= {}^nC_1(2 - 1) + {}^nC_2.(2^2 - 1) + \ldots + {}^nC_n(2^n - 1)$$

$$= {2 + 1}^n - 1 + {1 + 1}^n - 1 = 3^n - 2^n$$

33. Though it may appear that this problem is dependent on logarithmic manipulation but sych is not the case. We will prove it for general value $$z$$ rather than $$10$$

$$\sum_{r=0}^n{(-1)^r\binom{n}{r}\frac{1+r\log z}{(1+\log z^n)^r}}$$

$$=\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{\left(1+\log z^n\right)^r} +\sum_{r=1}^n(-1)^r\binom{n-1}{r-1}\frac{n\log z}{\left(1+\log z^n\right)^r}$$

$$=1+\sum_{r=1}^{n-1}(-1)^r\left(\binom{n-1}{r}+\binom{n-1}{r-1}\right)\frac{1}{\left(1+\log z^n\right)^r} +(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}$$

$$=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r} -\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r+1}} +(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}$$

$$=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r}+(-1)^n\frac{1}{(1+\log z^n)^r} -\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r}}+(-1)^n\frac{\log z^n}{(1+\log z^n)^n}$$

$$=1+(-1)^{n-1}\frac{1}{(1+\log z^n)^{n-1}}+(-1)^n\frac{1}{(1+\log z^n)^{n-1}}-1+(-1)^n\frac{\log z}{(1+\log z^n)^n}$$

$${=0}$$

34. $$32^{32} = (2^5)^{32} = 2^{160} = {3 - 1}^{160} = 3m + 1, m \in N$$

$${32}^{32^{32}} = 32^{3m + 1} = 2^{15m + 5} = 2^{3{5m + 1}}.2^2$$

$$=8^{5m + 1}.4 = 32.8^{5m} = 32(1 + 7)^{5m}$$

$$=32(1 + 7k), k \in N = 4 + 28 + 7(32k) = 4 + 7r, r \in N$$

Thus, remainder is $$4$$ when divided by $$7$$

35. Let $$t = x - 3,$$ then $$x - 2 = 1 + t$$

$$\sum_{r=0}^{2n}a_r(x - 2)^r = \sum_{r=0}^{2n}b_r(x - r)^r$$

$$\Rightarrow \sum_{r=0}^{2n}a_r(1 + t)^r = \sum_{r=0}^{2n}b_rt^r$$

Equating coefficients of $$t^n$$ we obtain desired result.

36. Given expression is $$= (1 + x)^{1000}\left[1 + 2\frac{x}{1 + x} + 3\left(\frac{x}{1 + x}\right)^2 + \ldots + 1001\left(\frac{x}{1 + x}\right)^{1000}\right]$$

The series is arithmetico-geometric series. Solving it yields

$$= (1 + x)^1002 - x^{1001}(1 + x) - 1001.x^{1001}$$

Required coefficient of $$x^{50} = {}^{1000}C_50$$

37. L.H.S. = coeff. of $$x^n$$ in $$(1 + x)^n + (1 + x)^{n + 1} + \ldots + (1 + x)^{n + k}$$

$$(1 + x)^n + (1 + x)^{n + 1} + \ldots + (1 + x)^{n + k}$$

$$= (1 + x)^n\left[\frac{1 + x}^{k + 1 - 1}{x}\right]$$

$$= \frac{1}{x}(1 + x)^{n + k + 1} - \frac{1}{x}(1 + x)^n$$

Coeff. of $$x^n = {}^{n + k + 1}C_{n + 1}$$

Hence, we have proved the desired equation.

38. Let $$S = (x + 2x^2 + 3x^3 + \ldots + nx^n)$$

$$xS = x^2 + 2x^3 + \ldots + (n - 1)x^n + nx^{n + 1}$$

$$S = x\frac{1 - x^n}{(1 - x)^2} - n\frac{x^{n + 1}}{1 - x}$$

$$(1 + x + 2x^2 + 3x^3 + \ldots + nx^n)^2 = \left[1 + \frac{x(1 - x^n)}{1 - x}^2 - \frac{nx^{n + 1}}{1 - x}\right]^2$$

Required coeff. of $$x^n$$ = coeff. of $$x^n$$ in $$\left(1 + \frac{x}{(1 - x)^2}\right)^2$$ [leaving terms containing powers of $$x$$ greater than $$n$$]

$$=$$ coeff. of $$x^n$$ in $$\left[1 + \frac{2x}{(1 - x)^2} + \frac{x^2}{(1 - x)^4}\right]$$

Solving this yields answer as $$\frac{n(n ^2 + 11)}{6}$$

39. Let $$S = 1 + (1 + x) + (1 + x)^2 + \ldots + (1 + x)^n$$

$$(1 + x)S = (1 + x) + (1 + x)^2 + \ldots + (1 + x)^n + (1 + x)^{n + 1}$$

Subtracting, we get

$$xS = (1 + x)^{n + 1} - 1$$

$$\therefore$$ in $$S = {}^{n + 1}C_{k + 1}$$

40. Let the expression be $$E = (x + 1)^n + (x + 1)^{n - 1}(x + 2) + (x + 1)^{n - 2}(x + 2)^2 + \ldots + (x + 2)^n$$

We know that $$(x - y)(x^{n - 1} + x^{n - 2}y + \ldots + y) = x^n - y^n$$

Thus, $$E(x + 2 - x - 1) = (x + 2)^{n + 1} - (x + 1)^{n + 1}$$

Coeff. of $$x^3$$ in $$(x + 2)^{n + 1} - (x + 1)^{n + 1}$$ is

$$= {}^{n + 1}C_3.2^{n - 2} - {}^{n + 1}C_3$$

41. $$\left(\frac{a + 1}{a^{\frac{2}{3}} - a^{\frac{1}{3}} + 1} - \frac{a - 1}{a - a^{\frac{1}{2}}}\right)^{10}$$

$$= (\sqrt[3]{a} - \frac{1}{\sqrt{a}})^{10}$$

$$t_{r + 1} = {}^{10}C_r a^{\frac{10 - r}{3}}a^{-\frac{r}{2}}$$

Since the term has to be independent of $$a, \Rightarrow \frac{10 - r}{3} - \frac{r}{2} \Rightarrow r = 4$$

Thus $$t_5 = {}^{10}C_4 = 210$$

42. Coeff. of $$x^2$$ in $$\left(x + \frac{1}{x}\right)^{10}(1 - x + 2x^2) =$$ coeff. of $$x^2$$ in $$\left(x + \frac{1}{x}\right)^{10}$$ - coeff. of $$x$$ in $$\left(x + \frac{1}{x}\right)^{10}$$ - 2 * coeff. of term independent of $$x$$ in $$\left(x + \frac{1}{x}\right)^{10}$$

$$(r + 1)^{th}$$ term in $$\left(x + \frac{1}{x}\right)^{10} = {}^{10}C_rx^{10 - r}x^{-r} = {}^{10}C_rx^{10 - 2r}$$

Coeff. of $$x^2$$ means $$10 - 2r = 2 \Rightarrow r = 4.$$ Thus, coeff. $$= {}^10C_4 = 210$$

Coeff. of $$x$$ means $$10 - 2r = 1$$ which makes $$r$$ a fraction. Thus, coeff. $$= 0$$

Coeff. of term indepdent of $$x$$ means $$10 -2r = 0 \Rightarrow r = 5.$$ Thus, coeff. $$= 2.{}^10C_5 = 504$$

Thus, final coeff. $$= 210 + 504 = 714$$

43. Coeff. of $$x^4$$ in $$(1 + x - 2x^2)^6 =$$ coeff. of $$x^r$$ in $$(1 + x(1 - 2x))^6$$

Thus coefficient of $$x^4$$ will occur in $$3^{rd}$$ terms onward.

Adding coefficients we get $$-45$$ as our answer.

44. We have $$(1 + x + 2x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$

$$= (1 + x + 2x^3)\left[\left(\frac{3}{2}x^2\right)^9 - {}^9C_1\left(\frac{2}{2}x^2\right)^8\frac{1}{3x} + \ldots + (-1)^9\left(\frac{1}{3x}\right)^9\right]$$

Thus, the term independent of $$x$$ in the expansion is $$1a_0 + 1a_1 + 2a_3$$ where $$a_m$$ is the coefficient of $$x^m$$ in the second bracket $$[]$$ of previous equation. Now, $$(r + 1)^{th}$$ term in $$[]$$ of previous equation is

$${}^9C_r\left(\frac{3}{2}x^2\right)^{9 - r}\left(-\frac{1}{3x}\right)^r = (-1)^r{}^9C_r\left(\frac{3}{2}\right)^{9 - r}\left(\frac{1}{3^r}\right)x^{18 -r}$$

$$\therefore a_{18 - 3r} =$$ coeff. of $$x^{18 - 3r}$$

For $$a_0, 18 - 3r = 0 \Rightarrow r = 6 \Rightarrow a_0 = {}^9C_6 \frac{3^3}{2^3}\frac{1}{3^6} = \frac{7}{18}$$

For $$a_1, 18 - 3s = 1, \Rightarrow r = \frac{19}{3}$$ which is fractional. $$\therefore a_1 = 0$$

For $$a_3, 18 - 3r = 3 \Rightarrow r - 7 \Rightarrow a_3 = -{}^9C_7\left(\frac{3}{2}\right)^2\frac{1}{3^7} = -\frac{1}{27}$$

Thus, required term $$= 1.\frac{7}{18} + .0 + 2.\frac{-1}{27} = \frac{17}{52}$$

45. Given $$\left(x^2 + \frac{1}{x^3}\right)^7(2 - x)^{10}$$ and we have to find term independent of $$x$$ in this.

Coeff of term independent of $$x$$ to be found in $$\frac{1}{x^{21}}(x^5 + 1)^7(2 - x)^{10}$$ i.e. coeff. of $$x^{21}$$ in $$(x^5 + 1)^7(2 - x)^{10}$$

$$(x^5 + 1)^7$$ will have coeff. of powers of $$x$$ where powers will be $$35, 30, 25, 20, \ldots, 0$$ while $$(2 - x)^{10}$$ will have powers of $$1, 2, 3, \ldots, 10$$

Clearly, combinations of $$20$$ and $$1$$ and $$15$$ and $$6$$ satisfy our needs.

Thus, computing these terms leads to answer of $$-61600$$

46. We have to find term independent of $$x$$ in $$(1 + x + x^{-2} + x^{-3})^{10}$$

Coeff. of term independent of $$x$$ in $$\frac{1}{x^30}(1 + x)^{10}(1 + x^3)^{10}$$

$$(1 + x)^{10}$$ will have powers of $$x$$ in $$0, 1, 2, 3, \ldots, 10$$ while $$(1 + x^3)^{10}$$ will have powers of $$x$$ in $$0, 3, 6, 9, \ldots, 30$$

Thus, combinaions of $$(0, 30), (3, 27), (6, 24), (9, 21)$$ are the combinations which will satisfy our needs, where first number is power of $$x$$ in $$(1 + x)^{10}$$ and second number is power of $$x$$ in $$(1 + x^3)^{10}$$

Thus coeff. is $$= {}^{10}C_1.{}^{30}C_{30} + {}^{10}C_3.{}^{30}C_{27} + {}^{10}C_6.{}^{30}C_{24} + {}^{10}C_9.{}^{30}C_{21}$$

47. Given $$(1 + x^2)^2(1 + x)^n = \sum_{k = 0}^{n + 4}a_kx^k$$

$$a_1, a_2, a_3$$ are coefficients of $$x, x^2, x^3$$ respectively.

Thus, we can find $$a_1, a_2, a_3$$ like we did in 143 and then apply arithmetic progression confition $$2a_2 = a_1 + a_3$$ to find $$n$$

48. We have to prove that $${}^mC_1 + {}^{m + 1}C_2 + {}^{m + 2}C_3 + \ldots + {}^{m + n - 1}C_n = {}^nC_1 + {}^{n + 1}C_2 + {}^{n + 2}C_3 + \ldots + {}^{n + m - 1}C_n$$

Keep in mind that $${}^mC_0 = {}^nC_0$$

Adding $${}^mC_0$$ to L.H.S., we get $${}^mC_0 + {}^mC_1 + {}^{m + 1}C_2 + {}^{m + 2}C_3 + \ldots + {}^{m + n - 1}C_n$$

Now we know that $${}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r$$ applying the repeatedly we obtain L.H.S. as $${}^{m + n}C_n$$

Similarly adding $${}^nC_0$$ to R.H.S. and applying above formula repeatedly we obtain R.H.S. as $${}^{m + n}C_m$$

Clearly, $${}^{m + n}C_m = {}^{m + n}C_n$$

49. Let us solve these one by one.

1. We observe that $$1 + x + x^2 = (x + \omega)(x + \overline{\omega})$$ where $$\omega$$ and $$\overline{\omega}$$ are cube root of unity not equal to $$-1$$

$$\sum_{r=0}^{2n}c_rx^r=(x^2+x+1)^n=(x+w)^n(x+\overline{w})^n=\sum_{k=0}^n\binom{n}{k}x^kw^{n-k}\cdot \sum_{l=0}^n\binom{n}{\ell}x^l \overline{w}^{n-l}$$

Thus, $$a_r=\sum_{j=0}^r\binom{n}{j}\binom{n}{r-j}w^{n-j}\overline{w}^{\,n-(r-j)}=\sum_{j=0}^r\binom{n}{j}\binom{n}{r-j}w^{r-2j}$$

Thus, $$a_r = a_{2n - r}$$

1. $$a_r = a_{2n - r}$$ from previous part.

Substituting $$x = 1$$

$$a_0 + a_1 + \ldots + a_2n = 2(a_0) + 2(a _1) + \ldots + 2a_{n - 1} + a_n = 3^n$$

$$\Rightarrow a_0 + a_1 + a_2 + \ldots + a_{n - 1} = \frac{1}{2}(3^n - a_n)$$

2. Differentiating and using result obtained in part 1 we can prove this.

1. Given $$\frac{(1 - x^3)^n}{(1 - x)^{3n}} = \sum_{r = 0}^na_r\frac{x^r}{(1 - x)^{2r}}$$

$$\Rightarrow \left(\frac{1 + x + x^2}{(1 - x)^2}\right)^n = \sum_{r = 0}^na_r\alpha^r,$$ where $$\alpha = \frac{x}{(1 - x)^2}$$

$$(1 + 3\alpha)^n = \sum_{r = 0}^na_r\alpha^r$$

Coeff of $$\alpha^r = a_r = {}^nC_r3^r$$

2. Coeff. of middle term of $$(1+ x)^{2n} = {}^{2n}C_n$$

Coeff. of $$x^n$$ in $$(1 + x)^{2n - 1} = {}^{2n - 1}C_n$$

Clearly, $$2.{}^{2n - 1}C_n = {}^{2n}C_n$$

3. Coeff. of terms in the expansion of $$(x + y)^{200}$$ are $${}^{200}C_1, {}^{200}C_2, {}^{200}C_3, \ldots, {}^{200}C_{200}$$

Since middle term has greatest coefficient, therefore $$r = 100$$

4. Let committees of $$r$$ persons be made out of $$20$$ persons. Then number of committees $$= {}^20C_r$$

Since middle term has greatest coefficient therefore $${}^{20}C_r$$ will be maximum when $$r = 10$$

Therefore, $$10$$ persons should be chosen for maximum no. of committees.

5. This problem is similar to 149 and 150 and has been left as an exercise.

6. $$(3 + 2x)^7$$ will have $$8$$ terms with $$4^{th}$$ and $$5^{th}$$ terms as middle terms. We know that when there are two middle terms coefficients are equal. Therefore, these two middle terms are consecutive terms which have equal coefficient.

7. Let $$(1 + 5x^2 - 7x^3)^{2000} = a_0 + a_1x + a_2x + \ldots + a_{6000}x^{6000}$$

Substituting $$x = 1,$$ we get

$$(1 + 5 - 7)^{2000} =$$ sum of coefficients $$= 1$$

8. Substitutin $$x^{-\frac{x}{4}} = 1$$ and $$x^{\frac{5x}{4}} = 1$$

$$(1 + 1)^n = 64 \Rightarrow n = 6$$

Hence greatest term = middle term = $$4^{th}$$ term

According to question $$t_4 = (n - 1)+ t_3$$

Solving this gives us, $$[\alpha] = 0$$

9. Let $$S$$ be the sum of coeff. in $$(5p - 4q)^n$$. Substituting $$p =1, q = 1,$$ we get

$$S = (5 - 4)^n = 1$$

10. Let $$S$$ be the sum of coeff. in $$(5p - 4q)^n$$. Substituting $$x = 1,$$ we get

$$S = (1 - 3 + 1)^{201}(1 + 5 - 5)^{503} = -1$$

11. Substituing $$x =1$$ reduces the two expansions as $$(t - 1)^n$$ and $$(1 - t)^n$$

Clearly, if $$n$$ is odd then sign will differe except when $$t = 1.$$ However, if $$n$$ is even both expansions will be positive and $$t$$ can assume any real value.

12. Substituting $$x = 1, i, -i$$ and then multiplying when $$x = i, -i$$ we obtain the desired result.

13. Let $$r^{th}$$ term be the greatest term.

$$t_r = \sqrt{3}\left[{}^{20}C_{r - 1}\left(\frac{1}{\sqrt{3}}\right)^{r - 1}\right]$$

$$t_{r + 1} = \sqrt{3}\left[{}^{20}C_r\left(\frac{1}{\sqrt{3}}\right)^r\right]$$

$$\frac{t_r}{t_{r + 1}} = \frac{{}^{20}C_{r - 1}}{{}^{20}C_r}\sqrt{r}$$

$$= \frac{r}{21 - r}\sqrt{3} \geq 1 \Rightarrow r = 7.69$$

Also, $$\frac{t_{r -1}}{t_r} \leq 1$$ gives us $$r = 8.5$$

Hence, $$8^{th}$$ term will be greatest term and $$t_8 = \frac{25840}{9}$$

14. Since $$t_{11}$$ is G.M. of $$t_8$$ and $$t_{12}$$

$$t_{11} = \sqrt{t_8.t_{12}}$$

$$\left(\frac{15!}{10!5!}x^5a^{10}\right)^2 = \frac{15!15!}{7!8!11!4!}x^{12}a^{18}$$

$$\Rightarrow \frac{x}{a} = \sqrt{\frac{77}{75}}$$

Let $$r^{th}$$ term be the greatest term. Now,

$$t_r = {}^{15}C_{r - 1}x^{16 - r}a^{r - 1}$$

$$t_{r + 1} = {}^{15C_r}x^{15 - r}a^r$$

$$\frac{t_r}{t_{r + 1}} = \frac{r}{16 - r}\frac{x}{a} \geq 1$$

$$r \geq 7.947$$

Hence, $$8^{th}$$ term is the greatest term.

15. $$t_{n + 1} = {}^{2n}C_nx^n, t_{n + 2} = {}^{2n}C_{n + 1}x^{n + 1}, t_n = {}^{2n}C_{n - 1}x^{n - 1}$$

$$\frac{t_{n + 1}}}{t_{n + 2}} = \frac{n + 1}{n}.\frac{1}{x} > 1 \Rightarrow x < \frac{n + 1}{n}$$

Also, $$t_{n + 1} = \frac{n + 1}{n}x > 1 \Rightarrow x > \frac{n}{n + 1}$$

Thus, greatest term will have greatest coefficient if and only if $$x \in \left(\frac{n}{n + 1}, \frac{n + 1}{n}\right)$$

Given $$x \in \left(\frac{10}{11}, \frac{11}{10}\right)$$

Thus, $$n = 10$$

$$t_4 = \frac{n}{4} = \frac{5}{2}$$

$$\Rightarrow {}^{m}C_3(kx)^{m - 3}\frac{1}{x^3} = \frac{5}{2}$$

R.H.S. is independent of $$x, \Rightarrow m - 6 = 0, m = 6$$

$${}^{6}C_3k^3 = \frac{5}{2}\Rightarrow k = \frac{1}{2}$$

$$\Rightarrow mk = 3$$

16. This problem is similar to last problem and has been left as an exercise. The range is $$2 < x < \frac{64}{21}$$

17. $$9^n + 7 = (8 + 1)^n + 7 = {}^nC_0.8^n + {}^nC_18{n - 1} + \ldots + {}^nC_{n - 1}8 + {}^nC_n + 7$$

$$= 8k + 1 + 7, k \in N = 8(k + 1)$$

Thus, the number is divisible by $$8$$

18. Given expression is $$3^{2n + 1} + 2^{n + 2}$$ which can be rewritten as $$3.(7 + 2)^n + 4.2^n,$$ which upon expansion yields

$$3({}^nC_07^n + {}^nC_17^{n - 1}2 + \ldots + {}^nC_n2^n) + 4.2^n$$

$$= 7k + 2^n(3 + 4), k \in N$$

The above expression is divisible by $$7.$$

19. Let the binomial expansion be $$(x + y)^n$$ and $$a, b, c$$ be the coefficients of $$r^{th}, (r + 1)^{th}, (r + 2)^{th}$$ terms respectively.

Then, $$a = {}^nC_{r - 1}, b = {}^nC_r, c = {}^nC_{r + 1}$$

The descriminant of given quadratic equation is $$D = b^2 - 4ac$$

Substituting values of $$a, b, c$$ and simplifying we obtain

$$D = 4({}^nC_r)^2 \frac{n + 1}{(n - r + 1)(r + 1)}$$ where $$r$$ is non-negative integer.

Clearly $$D > 0,$$ hence roots of given equation are real and unequal.

20. This problem is similar to previous one and has been left as an exercise.

21. Calculus Method:

$$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Multiplying with $$x$$ and differentiating w.r.t. $$x,$$ we get

$$(1 + x)^n + nx(1 + x)^{n - 1} = C_0 + 2xC_1 + 3x^2C_2 + (n + 1)x^nC_n$$

Substituting $$x = -1,$$ we obtain

$$C_n - 2.C_1 + 3.C_2 - \ldots +(-1)^n(n + 1)C_n = 0$$

Second Method:

$$t_r = (-1)^{r - 1}r.{}^nC_{r - 1} = (-1)^{r - 1}(r- 1 + ).{}^nC_{r - 1} = (-1)^{r - 1}n.{}^{n - 1}C_{r - 2} + (-1)^{r - 1}.{}^nC_{r - 1}$$

$$\sum_{r = 1}^{n + 1} = -n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 - \ldots + {-1}^{n - 1}.{}^{n - 1}C_{n - 1}) + ({}^nC_0 - {}^nC_1 + \ldots + (-1)^n.{}^nC_n)$$

$$= -n(1 - 1)^{n - 1} + (1 - 1)^n = 0$$

22. Calculus Method:

We know that $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Substituting $$x = x^2$$ and multiplying with $$x,$$ we get

$$x(1 + x^2)^n = C_0x + C_1.x^3 + C_2.x^5 + \ldots + (2n + 1)x^{2n}.C_n$$

Differentiating both sides w.r.t $$x,$$ and substituting $$x = i,$$ we get

$$C_0 -3.C_1 + 5.C_2 - \ldots + (-1)^n(2n + 1)C_n = (1 - 1)^n + i.n.(1 - 1)^{n - 1}.2i = 0$$

Second Method:

$$t_r = = (-1)^{r - 1}(2r - 1).C_{r - 1} = (-1)^r[2(r - 1) + 1]{}^nC_{r - 1}$$

$$= 2(-1)^{r - 1}.n{}^{n - 1}C_{r - 1} + (-1)^{r - 1}{}^nC_{r - 1}$$

$$\sum_{r = 1}^{n + 1}t_r = -2n[{}^{n - 1}C_0 - {}^{n - 1}C_1 + \ldots + (-1)^{n - 1}{}^{n - 1}C_{n - 1}] + [{}^{n}C_0 - {}^nC_1 + \ldots + (-1)^n{}^nC_n]$$

$$= -2n(1 - 1)^{n - 1} + (1 - 1)^n = 0$$

23. Calculus Method:

L.H.S. = $$a[C_0 - C_1 + C_2 - \ldots + (-1)^n.C_n] + [1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n]$$

$$= a(1 - 1)^n + 1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n$$

Given, $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Differentiating w.r.t. $$x$$ and substuting $$x = -1,$$ we get

$$1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n = (1 - 1)^n = 0$$

Hence, desired equality is proved.

Second Method:

$$t_r = (-1)^{r - 1}[a - (r - 1)]{}^nC_{r -1}$$

$$= a(-1)^{r - 1}.{}^nC_{r - 1} - (-1)^{r - 1}n.{}^{n - 1}C_{r -1}$$

This can be proven to be $$0$$ like previous problems.

24. $$t_{r + 1} = r^r.{}^nC_rp^rq^{n - r}$$

$$= r.n.{}^{n - 1}C_{r - 1}p^rq^{n - r}$$

$$= n(r - 1 + 1){}^{n - 1}C_{r - 1}p^rq^{n - r}$$

$$= n[(n - 1).{}^{n - 2}C_{r - 2} + {}^{n - 1}C_{r - 1}]p^rq^{n - r}$$

L.H.S. $$= \sum_{r = 0}^n t_{r + 1}$$

$$= n(n - 1)p^2\sum_{r = 0}^n {}^{n - 2}C_{r - 2}q^{n -2 - (r - 2)} + np\sum_{r = 0}^{n}{}^{n - 1}C_{r - 1}p^{r - 1}q^{n - 1 - (r - 1)}$$

$$= n(n - 1)p^2(p + q)^{n - 2} + np(p + q)^{n - 1}$$

$$= n(n - 1)p^2 + np [\because p + q = 1]$$

$$= n^2p^2 + npq$$

25. $$(1 + x)^{10} = C_0 + C_1x + C_2x^2 + \ldots + C_10x^{10}$$

Integrating between limits $$0$$ and $$2$$ gives the desired result.

26. Calculus Method:

$$(1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n$$

$$\frac{1 - (1 - x)^n}{x} = {}^nC_1 - {}^nC_2x + \ldots + (-1)^{n - 1}C_{n}x^{n - 1} = \frac{x - (1 - x)^n}{x}$$

Integrating between limits $$0$$ and $$1,$$ we get

$$\left[{}^nC_1x - {}^nC_2\frac{x^2}{2} + \ldots + (-1)^{n - 1}\frac{x^n}{n}\right]_0^1 = \int_{0}^1\frac{1 - (1 - x)^n}{x}dx$$

Let $$z = 1 - x$$ then R.H.S. becomes $$\int_1^0 -\frac{1 - z^n}{1 - z}dz$$

$$= \int_0^1(1 + z + z^2 + \ldots + z^{n - 1})dz$$

$$= \left[z + \frac{z^2}{2} + \ldots + \frac{z^n}{n}\right]_0^1$$

$$= 1 + \frac{1}{2} +\frac{1}{3} + \ldots + \frac{1}{n}$$

Seccond Method:

Let $$S_n = C_1 - \frac{1}{2}C_2 + \frac{1}{3}C_3 - \ldots + \frac{(1)^nC_n}{n}$$

$$S_n = n - \frac{1}{2}\frac{n(n - 1)}{2!} + \frac{1}{3}\frac{n(n - 1)(n - 2)}{3!} + \ldots$$

$$= (n - 1 + 1) - \frac{1}{2}\frac{(n - 1)(n - 2 + 2)}{2!} + \frac{(n - 1)(n - 2)(n - 3 + 3)}{3!} + \ldots$$

$$= S_{n - 1} + \frac{1}{n}\left[n - \frac{n(n - 1)}{2!} + \frac{n(n - 1)(n - 2)}{3!} + \ldots\right]$$

$$= S_{n - 1} - \frac{1}{n}[C_0 - C1 + C_2 + \ldots -1] = S_{n - 1} - \frac{1}{n}[(1 - 1)^n - 1]$$

$$S_{n - 1} + \frac{1}{n} \therefore S_n - S_{n - 1} = 1$$

Similarly $$S_{n - 1} - S_{n - 2} = \frac{1}{n - 2}$$

$$S_{n - 2} - S_{n - 3} = \frac{1}{n - 3}$$

$$\ldots$$

$$S_2 - S_1 = \frac{1}{3}$$

$$S_1 = 1$$

Adding we get $$S_n = 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{n}$$

27. $$(1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n$$

Putting $$x = x^4,$$ and integrating between limits $$0$$ and $$1,$$ we get

$$C_0 - \frac{C_1}{5} + \frac{C_2}{9} - \ldots + (-1)^n\frac{C_n}{4n + 1} = \int_{0}^1(1 - x^4)^ndx$$

It can be proven that $$\int_0^1(1 - x^4)^ndx = \frac{4n.n!}{1.5.9\ldots (4n + 1)}$$ [Refer to any book on calculus]

28. $$(1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n$$

Multiplying both sides with $$x^{n - 1},$$ and integrating with limit $$0$$ and $$1,$$ we get

$$\frac{C_0}{n} - \frac{C_1}{n + 1} + \frac{C_2}{n + 2} - \ldots + (-1)^n\frac{C_n}{2n} = \int_0^1 x^{n - 1}(1 - x)^ndx$$

Integrating R.H.S. by parts we obtain the desired result.

29. L.H.S. $$= \left(\frac{C_0}{n} - \frac{C_0}{n + 1}\right) - \left(\frac{C_1}{n + 1} - \frac{C_1}{n + 2}\right) + \ldots + (-1)^{n - 1}\left(\frac{C_n}{2n} - \frac{C_n}{2n + 1}\right)$$

$$= \left(\frac{C_0}{n} - \frac{C_1}{n + 1} + \frac{C_2}{n + 2} - \ldots + (-1)^n\frac{C_n}{2n}\right) - \left(\frac{C_0}{n + 1} - \frac{C_1}{n + 2} + \ldots + (-1)^n\frac{C_n}{2n + 1}\right)$$

Now this can be solved like 174.

30. This problem is similar to 174 and has been left as an exercise.

31. $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

$$(1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n$$

Multiplying these two we get

$$(C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n) = (1 - x^2)^n$$

Coeff. of $$x^n$$ on L.H.S. $$= C_0^2 - C_1^2 + C_2^2 - \ldots + (-1)^nC_n^2$$

R.H.S. $$= C_0 + (-1)^1C_1x^2 + C_2x^2 + \ldots$$

$$=$$ an expression having even powers of $$x$$

Thus, if $$n$$ is odd, coeff. of $$x^n$$ on R.H.S $$= 0$$

If $$n$$ is even, coeff. of $$x^n$$ on R.H.S. $$= (-1)^{\frac{n}{2}}{}^nC_{\frac{n}{2}}$$

$$= \frac{n!}{\left(\frac{n!}{2}\right)^2}$$

32. :math:(1 + x)^n = C_0 + C_1x + C_2x^2 + ldots + C_nx^n`X

$$(1 + x)^m = C_0 + C_1x + C_2x^2 + \ldots + C_mx^m$$

Multiplying these two, we get

$$(C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0 + C_1x + C_2x^2 + \ldots + C_mx^m) = (1 + x)^{m + n} = {}^{m + n}C_0 + {}^{m + n}C_1x + \ldots + {}^{m + n}C_rx^r + \ldots + {}^{m + n}C_{m + n}x^{m + n}$$

Equating the coefficient of $$x^r$$ we get the desired result.

33. $$(1 + x)^{2n} = C_0 + C_1x + C_2x^2 + \ldots + C_{2n}x^{2n}$$

$$(x + 1)^{2n} = C_0x^{2n} + C_1x^{2n - 1} + C_2x^{2n - 2} + \ldots + C_2n$$

Multiplying we get

$$(C_0 + C_1x + C_2x^2 + \ldots + C_{2n}x^{2n})(C_0x^{2n} + C_1x^{2n - 1} + C_2x^{2n - 2} + \ldots + C_{2n}) = (1 - x^2)^{n}$$

Equating the coefficients of $$x^{2n}$$ we get the desired result.

34. $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Differentaiating w.r.t $$x,$$ we obtain

$$n(1 + x)^{n - 1} = C_1 + 2C_2x + \ldots + nC_nx^{n -1}$$

Also, $$(x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n$$

Multiplying two previous equations, we obtain

$$n(1 + x)^{2n - 1} = (C_1 + 2C_2x + \ldots + nC_nx^{n -1})(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)$$

Equating coefficients of $$x^{n - 1},$$ we obtain

$$C_1^2 + 2.C_2^2 + 2.C_3^3 + \ldots + n.C_n^2 = n.{}^{2n - 1}C_{n - 1} = \frac{(2n - 1)!}{[(n - 1)!]^2}$$

35. $$(1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n$$

Integrating with limits $$0$$ and $$x,$$ we obtain

$$\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^x = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^x$$

Also, $$(x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n$$

Multiplying and equating coefficients of $$x^{n + 1},$$ we obtain

$$C_0^2 + \frac{C_1^2}{2} + \frac{C_2^2}{3} + \ldots + \frac{C_n^2}{n + 1} = \frac{{}^{2n + 1}C_{n + 1} - 0}{n + 1} = \frac{(2n + 1)!}{[(n + 1)!]^2}$$

36. $$(1 - x)^n = C_0 - C_1x + C_2x^2 - \ldots + (-1)^nC_nx^n$$

Multiplying with $$x,$$ we obtain

$$x(1 - x)^n = C_0x + C_1x^2 + \ldots + (-1)^nC_nx^n$$

Differentiating w.r.t $$x,$$ we get

$$(1 - x)^n - nx(1 - x)^{n - 1} = C_0 -2xC_1 + \ldots + (-1)^n(n + 1)x^nC_n$$