Question

The value of $\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{9}+\sqrt{7}}+\cdots \cdots$ upto $100$ terms is

• A. $\frac{100}{\sqrt{203}+\sqrt{3}}$
• B. $\frac{100}{\sqrt{203}-\sqrt{3}}$
• C. $\frac{100}{\sqrt{201}-\sqrt{3}}$
• D. $\frac{100}{\sqrt{201}+\sqrt{3}}$

Answer 1

The correct option is A $\frac{100}{\sqrt{203}+\sqrt{3}}$

Let
$S_n=\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{9}+\sqrt{7}}+\cdots +\frac{1}{\sqrt{2n+3}+\sqrt{2n+1}}$
$\Rightarrow S_{100}=\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{9}+\sqrt{7}}+\cdots +\frac{1}{\sqrt{203}+\sqrt{201}}$

On rationalizing each term, we get
$S_{100}=\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{9}-\sqrt{7}}{2}+\cdots +\frac{\sqrt{203}-\sqrt{201}}{2}$
$=\frac{1}{2}(\sqrt{203}-\sqrt{3})$
$\therefore S_{100}=\frac{100}{\sqrt{203}+\sqrt{3}}$