Question

If $\frac{1}{1!10!}+\frac{1}{2!9!}+\frac{1}{3!8!}+.....+\frac{1}{1!10!}=\frac{2}{k!}(2^{k-1}-1)$ then find the value of k.

Answer 1

The given term can be simplified as

$2[\frac{1}{1!10!}+...\frac{1}{5!6!}]=\frac{2}{10!}[\frac{10!}{10!1!}+\frac{10!}{2!9!}+...]$

$\Rightarrow$ On further Simplification6

$\frac{2}{10!}[1+5+15+30+42]=3(2^{6-1}-1)\frac{2}{10!}$