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

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

Adding these two, we have

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

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

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 \(-1 < x \leq 1\)