Question

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

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

Answer 1

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

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

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

$f=\frac{1}{n!}{(}^n{C}_1{+}^n{C}_3+.....{.}^n{C}_{n-1})$

we k.n.t sum of coefficients of odd terms of $(1+x{)}^n$ is $2^{n-1}$.

$f=\frac{2^{n-1}}{n!}$