Question

The sum of the first $n$ terms of the series $\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+.....$ is

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

Answer 1

Answer: (A)

Let $S_n=\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}$ $+.....+\frac{1}{\sqrt{3n-1}+\sqrt{3n+2}}$
$=\frac{\sqrt{2}-\sqrt{5}}{-3}+\frac{\sqrt{5}-\sqrt{8}}{-3}+\frac{\sqrt{8}-\sqrt{11}}{-3}$ $+...+\frac{\sqrt{3n-1}-\sqrt{3n+2}}{-3}$
$=-\frac{1}{3}(\sqrt{2}-\sqrt{3n+2})$
$=\frac{1}{3}(\sqrt{3n+2}-\sqrt{2})$