Question

The sum of the series $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots$ upto $15$ terms is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

Given series is
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots \dots$
$n^{th}term=\frac{1}{\sqrt{n}+\sqrt{n+1}}$
$\therefore 15^{th}term=\frac{1}{\sqrt{15}+\sqrt{16}}$
Thus, given series upto 15 terms is
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots \dots +\frac{1}{\sqrt{15}-\sqrt{16}}$
This can be re-written as
$\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+......+\frac{\sqrt{15}-\sqrt{16}}{-1}$
(By rationalization)
$=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+....-\sqrt{14}+\sqrt{15}-\sqrt{15}+\sqrt{16}$
$=-1+\sqrt{16}=-1+4=3$
Hence, the required sum $=3$