Question

If $nϵN$ and $n$ is even, then $\frac{1}{1.(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+....+\frac{1}{(n-1)!1!}$ equals.

Answer 1

$S=\frac{1}{1(n-1)}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+......+\frac{1}{(n-1)!(n-1)!}$

$r^{th}$ term $=\frac{1}{(2r-1)!(n-(2r-1\left)\right)!}$

$S\times n!{=}^n{C}_1{+}^n{C}_3{+}^n{C}_5+.....{.}^n{C}_{n-1}$

$S_{even}\times n!{=}^n{C}_0{+}^n{C}_2{+}^n{C}_4+......{.}^n{C}_n.$

${}^n{C}_0{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3+........{.}^n{C}_n+2^n$ ........(1)

${}^n{C}_0{-}^n{C}_1{+}^n{C}_2{-}^n{C}_3+........{+}^n{C}_n=0$ ........(2)

(1)-(2)

$2{(}^n{C}_1{+}^n{C}_3{+}^n{C}_5+.......{.}^n{C}_{n=1})=2^n$

${}^n{C}_1{+}^n{C}_3+.......{+}^n{C}_{n-1}=2^n$

$s\times n!=2^n$

$\therefore S=\frac{2^n}{n!}$