Question

Series $\frac{1}{1!(n-1)}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+...$ is equal to:

• A. $\frac{2^n}{n!}$
• B. $\frac{2^{n-1}}{n!}$
• C. 0
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+....$
Multiplying numerator and denominator by $n!$
$=\frac{1}{n!}\left[\frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}+\frac{n!}{5!(n-5)!}+...\right]$
$=\frac{1}{n!}\left[{}^n{C}_1{+}^n{C}_3{+}^n{C}_5+...\right]$
$=\frac{1}{n!}{2}^{n-1}$