Question

The sum of the series ${\sum }_{k=1}^{360}\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}$ is

• A. $\frac{18}{19}$
• B. $\frac{19}{18}$
• C. $\frac{13}{19}$
• D. $\frac{14}{13}$

Answer 1

The correct option is A

$\frac{18}{19}$

$T_k=\frac{1}{\sqrt{k+1}\sqrt{k}(\sqrt{k}+\sqrt{k+1})}$

$=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$

$S={\sum }_{k=1}^{360}{T}_k$

$=1-\frac{1}{\sqrt{361}}=\frac{18}{19}$