Sum of series $\;S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\;....\;+\frac{1}{100\sqrt{99}+99\sqrt{100}} \;is$

Question

$(a)\;\frac{1}{10}\qquad(b)\;\frac{9}{10}\qquad(c)\;\frac{3}{10}\qquad(d)\;None$

Answer 1

Answer : $(b)\;\frac{9}{10}$

Explanation : $a_{k}=\frac{1}{\sqrt{k+1}\;k+\sqrt{k}\;(k+1)}$

$=\frac{1}{\sqrt{k(k+1)}}\;[\frac{1}{\sqrt{k}+\sqrt{k+1}}]$

$\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\;(k+1)}}$

$=\;[\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}]$

$S=\sum_{k=1}^{99}\;\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$

$=1-\frac{1}{\sqrt{100}}$

$=\frac{9}{10}.$