Question

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

Answer 1

Multiplying and dividing by $11!$ we get

$\left\{\frac{1}{11!}\right\}\{\frac{11!}{1!10!}+\frac{11!}{2!9!}+\frac{11!}{3!8!}+............+\frac{11!}{10!1!}\}$

$\Rightarrow \left\{\frac{1}{11!}\right\}\{{11}_{C_1}+{11}_{C_2}+{11}_{C_3....................}+{11}_{C_{10}}\}$

$\Rightarrow \left\{\frac{1}{11!}\right\}\{2^{11}-2\}=\left\{\frac{2}{11!}\right\}\{2^{10}-1\}$

by comparing we get $\Rightarrow k=11$