Question

In the expansion of $(1+a{)}^{m+n}$ prove that coefficients of $a^m$ and $a^n$ are equal.

Answer 1

It is known that $(r+1{)}^{th}$ term $\left(T_{r+1}\right)$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-1}{b}^r$

Assuming that $a^m$ occurs in the $(r+1{)}^{th}$ term of the expansion $(1+a{)}^{m+n}$ we obtain

$T_{r+1}{=}^{m+n}{C}_r\left(1{)}^{m+n-r}\right(a{)}^r{=}^{m+n}{C}_r{a}^r$

Comparing the indices of a in $a^m$ and in $T_{r+1}$ we obtain $r=m$

Therefore the coefficient of $a^m$ is
${}^{m+n}{C}_m=\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m+n)!}{m!n!}....\left(1\right)$

Assuming that $a^n$ occurs in the $(k+1)$th term of expansion $(1+a{)}^{m+n}$, we obtain

$T_{k+1}{=}^{m+n}{C}_k\left(1{)}^{m+n-k}\right(a{)}^k{=}^{m+n}{C}_k{a}^k$

Comparing the indices of a in $a^n$ and in $T_{k+1}$ we obtain $k=n$

Therefore the coefficient of $a^n$ is

${}^{m+n}{C}_n=\frac{(m+n)!}{n!(m+n-n)!}=\frac{(m+n)!}{n!m!}....\left(2\right)$

Thus from $1$ and $2$ it can be observed that the coefficients of $a^m$ and $a^n$ in the expansion of $(1+a{)}^{m+n}$ are equal.