Q. For any integer $n \geq 1$, the sum $\sum_{k = 1}^{n}k(k + 2)$ is equal to
A) $\frac{n(n + 1)(n + 2)}{6}$
B) $\frac{n(n + 1)(2n + 1)}{6}$
C) $\frac{n(n + 1)(2n + 7)}{6}$
D) $\frac{n(n + 1)(2n + 9)}{6}$
Explanation
(c)
$\sum_{k = 1}^{n}{\left( k^{2} + 2k \right) = \sum_{k = 1}^{n}{k^{2} + 2\sum_{k = 1}^{n}k}}$
$= \frac{n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}$
$= \frac{n(n + 1)(2n + 7)}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.