Question

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

• A. ; for all even values of n
• B. , for all values of n i.e., all even odd values
• C. 0
• D. None of these

Answer 1

The correct option is B

, for all values of n i.e., all even odd values

Multiplying each term by n! the question

reduces to

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

= ${}^n{C}_1$ + ${}^n{C}_3$ + ${}^n{C}_5$ + ........ = $2^{n-1}$.

Thus $\frac{1}{1!(n-1)!}$ + $\frac{1}{3!(n-3)!}$ + $\frac{1}{5!(n-5)!}$ +........= $\frac{1}{n!}$$2^{n-1}$.