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. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

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