Question

The value of $[\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+....+\frac{1}{\sqrt{16}+\sqrt{15}}]$ is equal to

• A. 9
• B. 27
• C. 81
• D. 3

Answer 1

The correct option is D 3

$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+...+\frac{1}{\sqrt{16}+\sqrt{15}}$

By rationalising the denominator of every term,

$=\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}\times \frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}-\sqrt{4}}+...+\frac{1}{\sqrt{16}+\sqrt{15}}\times \frac{\sqrt{16}-\sqrt{15}}{\sqrt{16}-\sqrt{15}}$

$=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+\frac{\sqrt{5}-\sqrt{4}}{5-4}+...+\frac{\sqrt{16}-\sqrt{15}}{16-15}$

$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{16}-\sqrt{15}$

$=-1+\sqrt{16}=-1+4=3$