Question

If $n\in N$ & $n$ is even, then $\frac{1}{1.\left(n-1\right)!}+\frac{1}{3!.\left(n-3\right)!}+\frac{1}{5!.\left(n-5\right)!}+......+\frac{1}{\left(n-1\right)!1!}=$

• A. $2^n$
• B. $\frac{2^{n+1}}{n!}$
• C. $2^{n}n!$
• D. none of these

Answer 1

The correct option is D none of these

Solution -

we know that,

${}^n{C}_0{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3+----{-}^n{C}_n=2^n$

Also

${}^n{C}_0{-}^n{C}_1{+}^n{C}_2{-}^n{C}_3+------{+}^n{C}_n=0$

Subtract -

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

$\frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}+--------\frac{n!}{(n-1)!1!}=2^{n-1}$

$\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+------\frac{1}{(n-1)!1!}=\frac{2^{n-1}}{n!}$

D is correct.