The sum of the series $ \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots…
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
1
2
3
4
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 $=3$