Question

If the co-efficients of the $(p+1{)}^{th}$ term and the $(p+3{)}^{th}$ term in the expansion of $(1+x{)}^{2n}$ be equal, then prove that $p=n-1$

Answer 1

$(1+x{)}^{2x}$

$T_{p+1}={}^{2x}{C}_p{x}^p$

$T_{p+3}={}^{2x}{C}_{p+2}{x}^{p+2}$

Coefficient of $T_{p+1}=$Coefficient $T_{p+3}$

$\Rightarrow {}^{2n}{C}_p={}^{2n}{C}_{p+2}$

$\Rightarrow \frac{\left(2n\right)!}{(2n-p)!p!}=\frac{\left(2n\right)!}{(2n-p-2)!(p+2)!}$

$\Rightarrow \frac{1}{(2n-p)(2n-p-1)}$

$=\frac{1}{(p+2)(p+1)}$

$\Rightarrow p^x+3p+2=4n^2-2np-2n-2np+p^z+p$

$\Rightarrow 4n^2-4np-2n-2p-2=0$

$\Rightarrow 2n^2-2np-n-p-1=0$

$\Rightarrow 2n^2-2np-n-p-1=0$

$\Rightarrow 2n^2-n-1=(2n+1)p$

$\Rightarrow p=\frac{2n^2-n-1}{2n+1}$

$p=\frac{(2n+1)(n-1)}{(2n+1)}$

$p=n-1$.