Question

If $\sum _{k=1}^{360}\left(\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}\right)=a,$ then the value of $19a$ is

Answer 1

Let $T_k=\frac{1}{\sqrt{k}\sqrt{k+1}[\sqrt{k}+\sqrt{k+1}]}$
$=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}$
$=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$
$\therefore S=\sum _{k=1}^{360}(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}})$
$=1-\frac{1}{\sqrt{361}}$
$=1-\frac{1}{19}=\frac{18}{19}=a\therefore 19a=18$