Question

The sum of $\frac{1}{\sqrt{2}+1}=\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{99}}$ is equal to

• A. 9
• B. 10
• C. 11
• D. 1
• E. None of these

Answer 1

The correct option is A

9

$\begin{array}{c}\frac{1}{\sqrt{2}+1}=\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{{\left(\sqrt{2}\right)}^2-1}=\sqrt{2}-1\\ =\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\sqrt{3}-\sqrt{2}\\ Similarly,\frac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3},......\frac{1}{\sqrt{100}+\sqrt{99}}=\sqrt{100}-\sqrt{99}\\ Byadding,weget\\ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{99}}=\sqrt{100}-1\\ =10-1=9.\end{array}$