Question

What is the value of $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}........$ upto 15 terms?

• A. 4
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is D

3

Rationalising each term, we get
$\frac{1}{1+\sqrt{2}}=\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{(\sqrt{2}{)}^2-1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1=\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}2-\sqrt{3}=\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\frac{1}{\sqrt{15}+\sqrt{16}}=\frac{1}{4+\sqrt{15}}\times \frac{4-\sqrt{15}}{4-\sqrt{15}}=\frac{4-\sqrt{15}}{16-15}=4-\sqrt{15}$
$\therefore$ Given expression
$=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(2-\sqrt{3})+...+4-\sqrt{15}=4-1=3$