Binomial Expression: An expression consisting of two terms, connected by $+$ or $-$ sign is called binomial expression.

Binomial Theorem: If $x$ and $a$ are real numbers then for all $n \in N$, we have

$(x+a)^n = ^{n} C_{0} \ x^n \ a^0 + ^{n} C_{1} \ x^{n-1} \ a^1 + ^{n} C_{2} \ x^{n-2} \ a^2 + \ldots \\ { \hspace{5.0cm} + ^{n} C_{r} \ x^{n-r} \ a^r + \ldots + ^{n} C_{n-1} \ x^1 \ a^{n-1} + ^{n} C_{n} \ x^0 \ a^n}$

$\Rightarrow (x+a)^n = \sum \limits_{r=0}^{n}$ $^{n} C_{r} \ x^{n-r} \ a^r$

This expression has the following properties:

i) It has $(n+1)$ terms

ii) The sum of the indices of $x$ and $a$ in each terms is n

iii) The coefficients of terms equidistant from the beginning and the end are equal.

iv) The general term $T_{r+1} = ^{n} C_{r} \ x^{n-r} \ a^r$

v) $(x+a)^n = \sum \limits_{r=0}^{n}$ $^{n} C_{r} \ x^{n-r} \ a^r$ can also be expressed as

$\displaystyle (x+a)^n = \sum \limits_{r+s=n}^{} \frac{n!}{r! s!} x^r \ a^s$

vi) Replacing $a$ by $-a$ in the expression $(x+a)^n$, we get

$(x-a)^n = ^{n} C_{0} \ x^n \ a^0 - ^{n} C_{1} \ x^{n-1} \ a^1 + ^{n} C_{2} \ x^{n-2} \ a^2 - ^{n} C_{3} \ x^{n-3} \ a^3+ \ldots \\ { \hspace{5.0cm} +(-1)^r {\ ^{n} C_{r} \ x^{n-r} \ a^r} + \ldots + (-1)^n {\ ^{n} C_{n} \ x^0 \ a^{n}} }$

The general term in the expansion of $(x-a)^n$ is $T_{r+1} = (-1)^n {\ ^{n} C_{r} \ x^{n-r} \ a^r }$

vii) Putting $x = 1$ and replacing $a$ by $x$,  in the expression $( x + a)^n$ we get

$(1+x)^n = ^{n} C_{0} + ^{n} C_{1} x + ^{n} C_{2} x^2+ \ldots + ^{n} C_{n} x^n = \sum \limits_{r=0}^{n} {^{n} C_{r} \ x^r}$

viii) Putting $a = 1$ in the expression $( x+a)^n$, we get

$(1+x)^n = ^{n} C_{0} x^n + ^{n} C_{1} x^{n-1} + ^{n} C_{2} x^{n-2}+ \ldots + ^{n} C_{n} x^0 = \sum \limits_{r=0}^{n} {^{n} C_{r} \ x^{n-r} }$

This is the expansion of $(1+x)^n$ in descending powers of $x$. In this case, $T_{r+1} = {\ ^{n} C_{r} \ x^{n-r} }$

$( x+a)^n + ( x-a)^n = 2 \Big\{ ^{n} C_{0} \ x^n \ a^0 + ^{n} C_{2} \ x^{n-2} \ a^2 + \ldots \Big\}$

$= 2 \{ \text{sum of the odd terms in the expression} (x+a)^n \}$

Similarly,

$( x+a)^n - ( x-a)^n = 2 \Big\{ ^{n} C_{1} \ x^{n-1} \ a^1 + ^{n} C_{3} \ x^{n-3} \ a^3 + \ldots \Big\}$

$= 2 \{ \text{sum of the even terms in the expression} (x+a)^n \}$

$\displaystyle \text{If } n \text{ is odd, then } \{ ( x+a)^n + ( x-a)^n \} \text{ and } \{ ( x+a)^n - ( x-a)^n \} \text{ both have } \Big( \frac{n+1}{2} \Big) \text{ terms. }$

$\displaystyle \text{If } n \text{ is even, then } \{ ( x+a)^n + ( x-a)^n \} \text{ has } \Big( \frac{n}{2} +1 \Big) \text{ terms whereas } \\ \\ \{ ( x+a)^n - ( x-a)^n \} \text{ both have } \Big( \frac{n}{2} \Big) \text{ terms. }$

x) If $O$ and $E$ denote respectively the sum of odd terms and even terms in the expansion of $(x+a)^n$, then

(a) $(x+a)^n = O+E$ and $(x-a)^n = O-E$

(b) $(x^2 - a^2)^n = O^2 - E^2$

(c) $4 OE = ( x - a)^{2n} - (x-a)^{2n}$

(d) $(x+a)^{2n} + ( x - a)^{2n} = 2 ( O^2 + E^2)$

$\displaystyle \text{xi) If } n \text{ is even, then} \Big( \frac{n}{2} + 1 \Big)^{th} \text{ term is the middle term. }$

$\displaystyle \text{If } n \text{ is off, then } \Big( \frac{n+1}{2} \Big)^{th} \text{ and } \Big( \frac{n+3}{2} \Big)^{th} \text{ are middle terms. }$