# 91. Binomials, Multinomials and Expansions

An algebraic expression containing one term is called a monomial, two terms is called binomial and more than two is called multinomial. Examples of a monimial expressions are $$2x, 4y,$$ examples of binomial expressions are $$a + b, x^2 + y^2, x^3 + y^3, x + \frac{1}{y}$$ and examples of multinomial expressions are $$1 + x + x^2, a^2 + 2ab + b^2, a^3 + 3a^2b + 3ab^2 + b^3.$$

## 91.1. Binomial Theorem

Newton gave this binomial theorem by which we can expand any power of a binomial expression as a series.

First we consider only positive integral values. For these the formula has following form:

$$(a + x)^n = {}^nC_0a^nx^0 + {}^nC_1a^{n - 1}x^1 + {}^nC_2a^{n - 2}x^2 + \ldots + {}^nC_na^0x^n$$

### 91.1.1. Proof by Mathematical Induction

Let $$P(n) = (a + x)^n = {}^nC_0a^nx^0 + {}^nC_1a^{n - 1}x^1 + {}^nC_2a^{n - 2}x^2 + \ldots + {}^nC_na^0x^n$$

When $$n = 1, P(1) = a + x = {}^1C_0a + {}^1C_1x$$

When $$n = 2, P(2) = a^2 + 2ax + x^2 = {}^2C_0a^2 + {}^2C_1ax + {}^2C_2x^2$$

Thus, we see that $$P(n)$$ holds good for $$n = 1$$ and $$n = 2$$

Let $$P(n)$$ is true for $$n = k$$ i.e.

$$P(k) = (a + x)^k = {}^kC_0a^kx^0 + {}^kC_1a^{k - 1}x^1 + {}^kC_2a^{k - 2}x^2 + \ldots + {}^kC_ka^0x^k$$

Multiplying both sides with $$(a + x)$$

$$P(k + 1) = (a + x)^{k + 1} = {}^kC_0a^{k + 1}x^0 + {}^kC_1a^kx + {}^kC_2a^{k - 1}x^2 + \ldots + {}^kC_ka^1x^k + \\{}^kC_0a^kx +{}^kC_1a^{k - 1}x^2 + {}^kC_2a^{k - 2}x^3 + \ldots + {}^kC_kx^{k + 1}$$

Combining terms with equal powers for $$a$$ and $$x,$$ using the formula $${}^nC_r + {}^nC_{r + 1} = {}^{n + 1}C_{r + 1}$$ and rewriting $${}^kC_0$$ and $${}^kC_k$$ as $${}^{k + 1}C_0$$ and $${}^{k + 1}C_{k + 1},$$ we get

$$P(k + 1) = (a + x)^{k + 1} = {}^{k + 1}C_0a^{k + 1}x^0 + {}^{k + 1}C_1a^kx^1 + {}^{k + 1}C_2a^{k - 1}x^2 + \ldots + {}^{k + 1}C_{k + 1}a^0x^{k + 1}$$

Thus, we see that $$P(n)$$ holds good for $$n = k + 1$$ and we have proven binomial theorem by mathematical induction.

### 91.1.2. Proof by Combination

We know that $$(a + x)^2 = (a + x)(a + x)\ldots$$ [$$n$$ factors]

If we see only $$a$$ then we see that $$a^n$$ exists and hence $$a^n$$ is a term in the final product. This is the term $$a^n$$ which can be written as $${}^nC_0a^nx^0$$

If we take the letter $$a, n - 1$$ times and $$x$$ once then we observe that $$x$$ can be taken ub $${}^nC_1$$ ways. Thus, we can say that the term in final product is $${}^nC_1a^{n - 1}x$$

Similarly, if we choose $$a, n - 2$$ times and $$x$$ twice then the term will be $${}^nC_2a^{n - 2}x^2$$

Finally, like $$a^n, x^n$$ will exist and can be written as $${}^nC_nx^n$$ for consistency.

Thus, we have proven binomial theorem by binomial induction.

## 91.2. Special Forms of Binomial Expansion

We have $$(a + x)^n = {}^nC_0a^nx^0 + {}^nC_1a^{n - 1}x^1 + {}^nC_2a^{n - 2}x^2 + \ldots + {}^nC_na^0x^n$$

1. Putting $$-x$$ instead of $$x$$

$$(a - x)^n = {}^nC_0a^nx^0 - {}^nC_1a^{n - 1}x^1 + {}^nC_2a^{n - 2}x^2 - \ldots + (-1)^n{}^nC_na^0x^n$$

2. Putting $$a = 1$$ in original equation

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

3. Puttin $$x = -x$$ in above case

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

## 91.3. General Term of a Binomial Expansion

We see that first terms is $$t_1 = {}^nC_0a^nx^0$$

Second term is $$t_2 = {}^nC_1a^{n - 1}x^1$$

So general term will be $$t_r = {}^nC_{r - 1}a^{n + 1 - r}x^{r - 1}$$

## 91.4. Middle Term of a Binomial Expansion

Case I: When $$n$$ is an even number.

Let $$n = 2m$$

Middle term will be $$m + 1$$ th term i.e. $${}^nC_ma^mx^m$$

Case II: When $$n$$ is an odd number.

Let $$n = 2m + 1$$ and in this case there will be two middle terms i.e. $$m + 1$$ th and $$m + 2$$ th terms will be middle terms.

So, $${}^nC_{m}a{m + 1}x^m$$ and $${}^nC_{m + 1}a^{m + 1}x^m$$ will be middle terms.

## 91.5. Largest Coefficient in a Binomial Expansion

In any binomial expansion middle term(s) have the largest coefficient. If there are two middle terms then their coefficients are equal.

## 91.6. Coefficients Equidistant from Start and End

Coefficints equidistant from start and end are equal.

Proof:

Coefficient of first term from start is $${}^nC_0$$

Coefficient of second term from start is $${}^nC_1$$

$$\ldots$$

Coefficient of rth term from start is $${}^nC_{r - 1}$$

Coefficient of first term end is $${}^nC_n$$

Coefficient of second term end is $${}^nC_{n - 1}$$

$$\ldots$$

Coefficient of rth term from end is $${}^nC_{n - r + 1}$$

We know that $${}^nC_{r - 1} = {}^nC_{n - r + 1}$$

Thus, coefficints equidistant from start and end are equal.

## 91.7. Properties of Binomial Coefficients

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

Putting $$x = 1,$$ we get

$$2^n = {}^nC_0 + {}^nC_1 + \ldots + {}^nC_n$$

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

$$0 = {}^nC_0 - {}^nC_1 + \ldots + (-1)^n{}^nC_n$$

$$2^n = 2[{}^nC_0 + {}^nC_2 + {}^nC_4 + \ldots]$$

$$2^{n - 1} = {}^nC_0 + {}^nC_2 + {}^nC_4 + \ldots$$

Subtracting, we get

$$2^{n - 1} = {}^nC_1 + {}^nC_3 + {}^nC_5 + \ldots$$

## 91.8. Multinomial Theorem

Consider the multinomial $$(x_1 + x_2 + \ldots + x_n)^p$$ where $$n$$ and $$p$$ are positive integers then if we can find the general term we can form the expansion.

The general term of such a multinomial is given by $$\frac{p!}{p_1!p_2!\ldots p_n!}x_1^{p1}x_2^{p2}x_3^{p3}\ldots x_n^{p_n}$$ such that $$p_1, p_2, \ldots p_n$$ are non-negative integers and $$p_1 + p_2 + p_n = p$$

Proof: We can find the general term using binomial theorem itself.

General term in the expansion $$[x_1 + (x_2 + x_3 + \ldots + x_n)]^n$$ is

$$\frac{n!}{p_1!(n - p_1)!}x_1^{p_1}(x_2 + x_3 + \ldots + x_n)^{n - p_1}$$

General term in expansion of $$(x_2 + x_3 + \ldots + x_n)^{n - p_1}$$ is

$$\frac{(n - p_1)!}{p_2!(n - p_1 - p_2)!}x_2^{p_2}(x_3 + x_4 + \ldots + x_n)^{n - p_1 - p_2}$$

Procedding thus, we obtain general term as

$$\frac{n!}{p_1!(n - p_1)!}\frac{(n - p_1)!}{p_2!(n - p_1 - p_2)!}\ldots x_1^{p_1}x_2^{p_2} \ldots x_n^{p_n}$$ where $$p_1 + p_2 + p_n = p$$ and $$p_1, p_2, \ldots p_n$$ are non-negative integers.

## 91.9. Some Results on Multinomial Expansions

1. No. of terms in the multinomial $$(x_1 + x_2 + \ldots + x_n)^p$$ is number of non-negative integral solution of the equation $$p_1 + p_2 + \ldots + p_n = p = {}^{n + p - 1}C_p$$ or $${}^{n + p - 1}C_{n - 1}$$

2. Largest coefficient in math:(x_1 + x_2 + ldots + x_n)^p is $$\frac{n!}{(q!)^{n - r}[(q + 1)!]^r}$$ where $$q$$ is the quotient and $$r$$ is the remainder of $$p/n$$

3. COefficient of $$x^r$$ in $$(a_0 + a_1x + a_2x^2 + \ldots + a_nx^n)^p$$ is $$\sum \frac{n!}{p_0!p_1!p_2!\ldots p_n!}a_0^{p_1} a_1^{p_2} a_n^{p_m}$$ where $$p_0, p_1, p_2, \ldots, p_n$$ are non-negative integers satisfying the equations $$p_0 + p_1 + \ldots + p_n = n$$ and $$p_1 + 2p_2 + \ldots + np_n = r$$

## 91.10. Binomial Theorem for any Index

### 91.10.1. To prove when index is a fractional quantity

Let $$f(m) = (1 + x)^m = 1 + mx + \frac{m(m - 1)}{1.2}x^2 + \frac{m(m - 1)(m - 2)}{1.2.3}x^3 + \ldots$$ where $$m \in R$$

then, $$f(n) = = (1 + x)^n = 1 + nx + \frac{n(n - 1)}{1.2}x^2 + \frac{n(n - 1)(n - 2)}{1.2.3}x^3 + \ldots$$

$$f(m)f(n) = (1 + x)^{m + n} = f(m + n)$$

$$f(m)f(n)\ldots$$ to $$k$$ factors $$= f(m + n + \ldots )$$ to $$k$$ terms

Letm $$m, n, \ldots$$ each equal to $$\frac{j}{k}$$

$$\left[f\left(\frac{j}{k}\right)\right]^k = f(j)$$

but $$j$$ is a postive integer, $$f(j) = (1 + x)^j$$

$$\therefore (1 + x)^{\frac{j}{k}} = f\left(\frac{j}{k}\right)$$

$$\therefore (1 + x)^{\frac{j}{k}} = 1 + \frac{j}{k}x + \frac{\frac{h}{k}\left(\frac{j}{k} - 1\right)}{1.2}x^2 + \ldots$$

And thus we have proven binomial theorem for any index.

### 91.10.2. To prove when index is a negative quantity

$$f(-n)f(n) = f(0) = 1$$

$$f(-n) = \frac{1}{f(n)} = (1 + x)^{-n} = 1 -nx + \frac{n(n + 1)}{1.2}x^2 + \ldots$$

## 91.11. General Term in Binomial Theorem of Any Index

General term is given by $$\frac{n(n - 1)\ldots(n - r + 1)}{r!}x^r$$

Note: The above exapansions do not hold true when $$x > 1$$ which can be quickly proven by making $$r$$ arbitrarily large.

For example, $$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \ldots$$

However, if we put $$x = 2,$$ then we have

$$(-1)^{-1} = 1 + 2 + 2^2 + 2^3 + \ldots$$ which shows that when $$x > 1$$ the above formula does not hold true.

From the chapter of G.P., we know that $$1 + x + x^2 + \dots$$ for $$r$$ terms is $$\frac{1}{1 - x} - \frac{x^r}{1 - x}$$

Thus, if $$r$$ is very large and $$x < 1$$ then we can ignore the second fraction but not when $$x > 1.$$

## 91.12. General Term of $$(1 - x)^{-n}$$

The $$(r + 1)^{th}$$ term is given by $$\frac{-n(-n - 1)\ldots (-n - r + 1)}{r!}(-x)^r$$

$$= \frac{n(n + 1)\ldots(n+ r - 1)}{r!}x^r$$

## 91.13. Exponential and Logarithmic Series

1. $$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots~\text{to}~\infty$$ where $$x$$ is any number. $$e$$ lies between $$2$$ and $$3$$. It also base for log natural which is written as $$\ln$$

2. $$e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots~\text{to}~\infty$$ where $$x$$ is any number

3. If $$a > 0, a^x = e^{x\log_e a} = 1 + \frac{x\log_ea}{1!} + \frac{(x\log_ea)^2}{2!} + \ldots~\text{to}~\infty$$

4. $$\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots~\text{to}~\infty$$ whre $$-1 < x \leq 1$$