Question

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

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

Answer 1

The correct option is D

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

Explanation for the correct option:.

Find the required sum.

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

So, the sum of $n$ terms of the given series is,

$$\begin{gathered} \frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}.....+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}} \\ =\frac{\sqrt{3}-\sqrt{1}}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+....+\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\left(\sqrt{2n+1}+\sqrt{2n-1}\right)\left(\sqrt{2n+1}-\sqrt{2n-1}\right)} \\ =\frac{\sqrt{3}-\sqrt{1}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{1}\right)}^2}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}+....+\frac{\sqrt{2n+1}-\sqrt{2n-1}}{{\left(\sqrt{2n+1}\right)}^2-{\left(\sqrt{2n-1}\right)}^2} \\ =\frac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+....+\sqrt{2n+1}-\sqrt{2n-1}\right) \\ =\frac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+....+\sqrt{2n+1}-\sqrt{2n-1}\right) \\ =\frac{1}{2}\left(\sqrt{2n+1}-\sqrt{1}\right) \\ =\frac{1}{2}\left(\sqrt{2n+1}-1\right) \end{gathered}$$

Hence, option (D) is the correct answer.