Q. The sum of the series $1 \bullet n + 1 + 2 \bullet (n - 1) + 3 \bullet (n - 2)\ldots + n \bullet 1$ is
A) $\frac{n(n + 1)(n + 2)}{6}$
B) $\frac{n(n + 1)(n + 2)}{3}$
C) $\frac{n(n + 1)(3n + 2)}{6}$
D) $\frac{n(n + 1)(2n + 2)}{3}$
Explanation
(a)
$1 \bullet n + 2(n - 1) + 3(n - 2) + ... + n \bullet 1$
$= \sum_{r = 1}^{n}{(n + 1)r - \sum_{r = 1}^{n}r^{2}}$
$= (n + 1)\sum_{}^{}n - \sum_{}^{}n^{2}$
$= \frac{(n + 1)n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6}$
$= \frac{n(n + 1)}{6}\left\{ 3n + 3 - 2n - 1 \right\}$
$= \frac{n(n + 1)(n + 2)}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.