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

Putting \(-x\) instead of \(x\)

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

Putting \(a = 1\) in original equation

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

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

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 \({}^{n + p - 1}C_{n - 1}\)

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 \(p/n\)

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

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

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

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

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