Question

Sum of the series
$P=\frac{1}{2\sqrt{1}+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is

• A. $1/10$
• B. $3/10$
• C. $9/10$
• D. $1/2$

Answer 1

The correct option is C $9/10$

$\text{Solve:}$

$\text{Given,}$

$p=\frac{1}{2\sqrt{1}+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\dots +\frac{1}{100\sqrt{99}+99\sqrt{10}}$

$\text{on rationalising the denominator of}$

$\text{each term we get,}$

$P=\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-3\sqrt{4}}{12}+\frac{5\sqrt{4}-4\sqrt{5}}{20}+\cdots \frac{100\sqrt{99}-99\sqrt{100}}{9900}$

$\Rightarrow P=1-\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{6}-\frac{2\sqrt{3}}{6}+\frac{4\sqrt{3}}{12}-\frac{3\sqrt{4}}{12}+\frac{5\sqrt{4}}{20}-\frac{4\sqrt{5}}{20}+-\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}$

$\Rightarrow P=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+\cdots \cdot \frac{\sqrt{99}}{99}-\frac{1}{\sqrt{100}}$

$\Rightarrow P=1-\frac{1}{\sqrt{{10}^8}}=1-\frac{1}{10}$

$\Rightarrow P=\frac{9}{10}$