Q. $The\ sum\ of\ the\ series\ 1 + \frac{1}{5} + \frac{1 \bullet 3}{5 \bullet 10} + \frac{1 \bullet 3 \bullet 5}{5 \bullet 10 \bullet 15} + .\ldots is$ equal to

A) $\frac{1}{\sqrt{5}}$
B) $\frac{1}{\sqrt{2}}$
C) $\sqrt{3}$
D) $\sqrt{\frac{5}{3}}$
Explanation

(d)
$Let\ S = 1 + \frac{1}{5} + \frac{1 \bullet 3}{5 \bullet 10} + \frac{1 \bullet 3 \bullet 5}{5 \bullet 10 \bullet 15} + .\ldots$
On comparing with
$(1 + x)^{n} = 1 + \frac{nx}{1!} + \frac{n(n - 1)}{2!}x^{2}$
$+ \frac{n(n - 1)(n - 2)}{3!}x^{3} + \ldots,we\ get$
$\Longrightarrow nx = \frac{1}{5}\ and\frac{n(n - 1)x^{2}}{2!} = \frac{1 \bullet 3}{5 \bullet 10}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ n = - \frac{1}{2}$
$and\ \ \ \ \ \ \ \ \ \ \ x = - \frac{2}{5}$
$\therefore sum = \left( 1 - \frac{2}{5} \right)^{- 1/2} = \left( \frac{3}{5} \right)^{- 1/2} = \sqrt{\frac{5}{3}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.