The average of the square of the number 0, 1, 2, 3, 4.........n is:

A) \( \Large \frac{1}{2}n \left(n+1\right) \)
B) \( \Large \frac{1}{6}n \left(2n+1\right) \)

C) \( \Large \frac{1}{6} \left(n+1\right) \left(2n+1\right) \)
D) \( \Large \frac{1}{6}n \left(n+1\right) \)

Correct answer:
B) \( \Large \frac{1}{6}n \left(2n+1\right) \)

Description for Correct answer:
Mean: \( \Large \frac{0^{2}+1^{2}+2^{2}+3^{2}+....+n^{2}}{ \left(n+1\right) } \)

= \( \Large \frac{n \left(n+1\right) \left(2n+1\right) }{6 \left(n+1\right) } = \frac{1}{6}n \left(2n+1\right) \)

Please provide the error details in above question

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