Question

The sum of the series $\sum _{k=1}^{62}\frac{1}{(k+2)\sqrt{k+1}+(k+1)\sqrt{k+2}}$.

Answer 1

${\sum }_{k=1}^{62}\frac{1}{(k+2)\sqrt{k+1}+(k+1)\sqrt{k+2}}$

${\sum }_{k=1}^{62}\frac{(k+2)-(k+1)}{(k+2)\sqrt{k+1}+(k+1)\sqrt{k+2}}$

$S_n={\sum }_{k=1}^{62}[\frac{1}{\sqrt{k+1}}-\frac{1}{\sqrt{k+2}}]$

$S_1=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}$

$S_2=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}$

$S_3=\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}$

$S_{62}=\frac{1}{\sqrt{63}}-\frac{1}{\sqrt{64}}$

Adding the above sums, we get

$S_{62}={\sum }_{k=1}^{62}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{64}}=\frac{1}{\sqrt{2}}-\frac{1}{8}=\frac{8-\sqrt{2}}{8\sqrt{2}}=\frac{8\sqrt{2}-2}{8.2}$

$=\frac{2(4\sqrt{2}-1)}{8\times 2}=\frac{4\sqrt{2}-1}{8}=S_{62}$