Question

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

• A. $\frac{2^{n-4}}{n!}$for even values of n only
• B. $\left(\frac{2^{n-4}+1}{n!}\right)-1$ for even values of n only
• C. $\frac{1}{n!}\left(2^{n-1}\right)$ for all values of n
• D. None of these

Answer 1

The correct option is C

$\frac{1}{n!}\left(2^{n-1}\right)$ for all values of n

Explanation for the correct option:

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

Multiply and divide the given series by $n!$

$\Rightarrow S=\frac{1}{n!}\left[\frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}+\frac{n!}{5!(n-5)!}\dots \dots \right]$

$\Rightarrow S=\frac{1}{n!}\left[{}^{n}C_1+{}^{n}C_3+{}^{n}C_5+\dots \dots .\right]$ $\left[\because {}^{n}C_r=\frac{n!}{(n-r)!}\right]$

$\Rightarrow S=\frac{1}{n!}\left[2^{n-1}\right]$ $\left[\because {}^{n}C_1+{}^{n}C_3+{}^{n}C_5+\dots ..=2^{n-1}\right]$

Hence, Option ‘C’ is Correct.