Question

The value of $\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\dots +\frac{1}{(n-1)!\left(1\right)!}$, where $n$ is even positive integer, is

• A. $\frac{2^n}{(n-1)!}$
• B. $\frac{2^{n-1}}{n!}$
• C. $\frac{2^{n+1}}{n!}$
• D. $\frac{2^n}{(n+1)!}$

Answer 1

The correct option is B $\frac{2^{n-1}}{n!}$

Let
$S=\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\dots +\frac{1}{(n-1)!\left(1\right)!}\Rightarrow S=\frac{1}{n!}[\frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}+\dots +\frac{n!}{(n-1)!\left(1\right)!}]\Rightarrow S=\frac{1}{n!}[{}^n{C}_1+{}^n{C}_3+\dots +{}^n{C}_{n-1}]\therefore S=\frac{2^{n-1}}{n!}$