Q41 / 90 · JEE Main 2012 (12 May Online) · Mathematics
The sum of the series
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots
$$
upto 15 terms is
- A1
- B2
- Correct answerC3
- D4
Solution
Given series is
$$
\begin{aligned}
& \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots . . \\
& n^{\text {th }} \text { term }=\frac{1}{\sqrt{n}+\sqrt{n+1}} \\
& \therefore 15^{\text {th }} \text { term }=\frac{1}{\sqrt{15}+\sqrt{16}}
\end{aligned}
$$
Thus, given series upto 15 terms is
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots+\frac{1}{\sqrt{15}+\sqrt{16}}
$$
This can be re-written as
$$
\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+\ldots \ldots+\frac{\sqrt{15}-\sqrt{16}}{-1}
$$
(By rationalization)
$$
\begin{gathered}
=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+\ldots . .-\sqrt{14}+\sqrt{15} \\
-\sqrt{15}+\sqrt{16} \\
=-1+\sqrt{16}=-1+4=3
\end{gathered}
$$
Hence, the required sum $=3$
Hence, the correct answer is option: 'C'.