Question

Show that the coefficient of $a^m$ and $a^n$ in the expansion of $(1+a{)}^{m+n}$ are equal.

Answer 1

General term of expansion $(a+b{)}^n$ is

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

For $(1+a{)}^{m+n}$

Putting $n=m+n,a=1,b=a$

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

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

For $a^m$, $a^r=a^m$

$\Rightarrow r=m$

$T_{m+1}{=}^{n+m}{C}_m(a{)}^m$

$=\frac{(n+m)!}{m!(m+m-m)!}(a{)}^m=\frac{(n+m)!}{m!\left(n\right)!}(a{)}^m$

For, $a^n$, $a^r=a^n$

$\Rightarrow r=n$

$Tn+1{=}^{m+m}{C}_n(a{)}^n$

$=\frac{(n+m)!}{n!\left(m\right)!}(a{)}^n$

Hence, the co-efficient of $a^m=$ co-efficient of $a^n$

$=\frac{(m+n)!}{m!n!}$