Q. The sum of $\frac{\frac{1}{2} \bullet \frac{2}{2}}{1^{3}} + \frac{\frac{2}{2}.\frac{3}{2}}{1^{3} + 2^{3}} + \frac{\frac{3}{2}.\frac{4}{2}}{1^{3} + 2^{3} + 3^{3}} + ...$ upto $n$ terms is equal to
A) $\frac{n - 1}{n}$
B) $\frac{n}{n + 1}$
C) $\frac{n + 1}{n + 2}$
D) $\frac{n + 1}{n}$
Explanation
(b)
The general term is
$T_{n} = \frac{\frac{n}{2}.\frac{n + 1}{2}}{1^{3} + 2^{3} + 3^{3} + ... + n^{3}}$
$= \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$
$\therefore\ S_{n} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.