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JEE MATHEMATICS MCQ: STATISTICS

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Understanding +4 | -1

The mean of the squares of first

n

natural numbers is

View answer and explanation

Correct option: D

\frac{1}{6}(n+1)(2 n+1)

.

Option D (

\frac{1}{6}(n+1)(2 n+1)

) is correct because the mean of the squares of the first n natural numbers is given by the formula

\frac{{\sum_{i=1}^{n} i^2}}{{n}} = \frac{{n(n+1)(2n+1)}}{{6n}} = \frac{{(n+1)(2n+1)}}{{6}}

.

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\bar X,

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