Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2}(\sqrt{2n+1}-1)$

Given,

$\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+................$

$\Rightarrow \frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}+\frac{\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{\left(\sqrt{5}-\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}+................$

$\Rightarrow \frac{1}{-2}\left\{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}+........\sqrt{2n+1}\right\}$

Therefore,

$=\frac{-1+\sqrt{2n+1}}{2}$

$=\frac{\sqrt{2n+1}-1}{2}$

Hence, this is the answer.