Question

The sum $S=\frac{1}{9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!}$ equals

• A. $\frac{2^9}{10!}$
• B. $\frac{2^{10}}{8!}$
• C. $\frac{2^{11}}{9!}$
• D. $\frac{2^{10}}{7!}$

Answer 1

The correct option is A $\frac{2^9}{10!}$

Let $S=\frac{1}{9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!}$
$=\frac{1}{10!}[\frac{10!}{9!}+\frac{10!}{3!7!}+\frac{10!}{5!5!}+\frac{10!}{7!3!}+\frac{10!}{9!}]$
$=\frac{1}{10!}[{}^{10}{C}_1+{}^{10}{C}_3+{}^{10}{C}_5+{}^{10}{C}_7+{}^{10}{C}_9]$
$=\frac{1}{10!}\left(2^{10-1}\right)$
$[\because C_1+C_3+C_5+....{.2}^{n-1}]$
$=\frac{2^9}{10!}$