Question

The sum of n terms of the series $\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...$ is

• A. $\sqrt{2n+3}$
• B. $\frac{\sqrt{2n+3}}{2}$
• C. $\sqrt{2n+3}-1$
• D. $\frac{\sqrt{2n+3}-\sqrt{3}}{2}$

Answer 1

Answer: (D)

$\frac{\sqrt{2n+3}-\sqrt{3}}{2}$

Let $S=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...\frac{1}{\sqrt{2n+1}+\sqrt{2n+3}}$
$\Rightarrow S=\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+\frac{\sqrt{7}-\sqrt{9}}{-2}+....+\frac{\sqrt{2n+1}-\sqrt{2n+3}}{-2}$
$\Rightarrow S=-\frac{1}{2}(\sqrt{3}-\sqrt{7}+\sqrt{5}-\sqrt{7}+....+\sqrt{2n+1}-\sqrt{2n+3})$
$\Rightarrow S=-\frac{1}{2}[\sqrt{3}-\sqrt{2n+3}]$
$\Rightarrow S=\frac{\sqrt{2n+3}-\sqrt{3}}{2}$
Ans: D