Q. 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) $\frac{1}{3}(\sqrt{2} - \sqrt{3n + 2})$
Explanation
(a)
Let $S_{n} = \frac{1}{\sqrt{2} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{8}} + ... + \frac{1}{\sqrt{3n - 1} + \sqrt{3n + 2}}$
$\frac{\sqrt{2} - \sqrt{5}}{- 3} + \frac{\sqrt{5} - \sqrt{8}}{- 3} + ... + \frac{\sqrt{3n - 1} - \sqrt{3n + 2}}{- 3}$
$= - \frac{1}{3}(\sqrt{2} - \sqrt{3n + 2}) = \frac{1}{3}(\sqrt{3n + 2} - \sqrt{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.