Question

If $\frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!1!}=\frac{2^n}{10!}$
Then n$=$

Answer 1

Given,

$\frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!1!}=\frac{2^n}{10!}$

$\frac{10!}{1!\cdot 9!}+\frac{10!}{3!\cdot 7!}+\frac{10!}{5!\cdot 5!}+\frac{10!}{7!\cdot 3!}+\frac{10!}{9!\cdot 1!}=2^n$

$\frac{20}{1!}+\frac{1440}{3!}+\frac{10!}{5!\cdot 5!}=2^n$

$\frac{20}{1!}+\frac{1440}{3!}+\frac{30240}{5!}=2^n$

$\frac{20}{1}+\frac{1440}{6}+\frac{30240}{120}=2^n$

$\frac{2400+28800+30240}{120}=2^n$

$512=2^n$

$2^9=2^n$

$\therefore n=9$