Question

If $n$ is an even integer,
$\frac{1}{1(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+\dots \dots \frac{1}{(n-1)!1}=$

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

Answer 1

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

Multiplying and dividing L.H.S with $n!$

$\frac{1}{n!}[\frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}....+\frac{n!}{(n-1)!1!}]$

we know that

${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\frac{1}{n!}{[}^n{C}_1+{}^n{C}_3+....+{}^n{C}_{n-1}]$

By formula

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

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