Q. The average of the squares of the numbers $0,\ 1,\ 2,\ 3,4,\ ...,\ n$ is
A) $\frac{1}{2}n(n + 1)$
B) $\frac{1}{6}n(2n + 1)$
C) $\frac{1}{6}(n + 1)(2n + 1)$
D) $\frac{1}{6}n(n + 1)$
Explanation
(b)
Mean $= \frac{0^{2} + 1^{2} + 2^{2} + 3^{2} + ... + n^{2}}{(n + 1)}$
$= \frac{n(n + 1)(2n + 1)}{6(n + 1)} = \frac{1}{6}n(2n + 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.