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

Let $S=\frac{1}{1!10!}+\frac{1}{2!9!}+\frac{1}{3!8!}+......+\frac{1}{10!1!}$

$\Rightarrow S\times 11!=\frac{11!}{1!10!}+\frac{11!}{2!9!}+\frac{11!}{3!8!}+......+\frac{11!}{10!1!}$

$\Rightarrow S\times 11!+2=\frac{11!}{0!11!}+\frac{11!}{1!10!}+\frac{11!}{2!9!}+\frac{11!}{3!8!}+......+\frac{11!}{10!1!}+\frac{11!}{11!0!}$

$\Rightarrow S\times 11!+2{=}^{11}{C}_0{+}^{11}{C}_1{+}^{11}{C}_2+.......{+}^{11}{C}_{11}$ --------- where ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\Rightarrow S\times 11!+2=2^{11}$ ------ (Using ${}^n{C}_0{+}^n{C}_1{+}^n{C}_2+.......{+}^n{C}_n=2^n$)

$\Rightarrow S=\frac{2}{11!}(2^{11-1}-1)$

Comparing this value of S with the expression in question, we get $k=2$.