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. $\frac{200}{\sqrt{218}+\sqrt{12}}$
• D. $\frac{200}{\sqrt{218}-\sqrt{12}}$

Answer 1

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

Let
$S_n=\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{9}+\sqrt{7}}+\cdots \cdots +T_n$
So $T_n=\frac{1}{\sqrt{2n+3}+\sqrt{2n+1}},n\in N$
$\therefore T_n=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2}=\frac{1}{2}[\sqrt{2n+3}-\sqrt{2n+1}]$
So,
$S_n=T_1+T_2+T_3+\cdots \cdots +T_n=\frac{1}{2}[\sqrt{5}-\sqrt{3}]+\frac{1}{2}[\sqrt{7}-\sqrt{5}]+\frac{1}{2}[\sqrt{9}-\sqrt{7}]+⋮⋮+\frac{1}{2}[\sqrt{2n+3}-\sqrt{2n+1}]S_n=\frac{1}{2}[\sqrt{2n+3}-\sqrt{3}]\Rightarrow S_n=\frac{n}{\sqrt{2n+3}+\sqrt{3}}\therefore S_{100}=\frac{100}{\sqrt{203}+\sqrt{3}}$