The mean of the squares of the first n natura | vedclass

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Question

(C) The first $n$ natural numbers are $1, 2, 3, \ldots, n$. Their squares are $1^2, 2^2, 3^2, \ldots, n^2$. $	ext{Mean} = \frac{	ext{Sum of observat

Vedclass · Accepted Answer

$\frac{2n^2+3n+1}{6}$

Vedclass · Answer

(C) The first $n$ natural numbers are $1, 2, 3, \ldots, n$. Their squares are $1^2, 2^2, 3^2, \ldots, n^2$. $	ext{Mean} = \frac{	ext{Sum of observations}}{	ext{Total number of observations}}$. $	ext{Mean} = \frac{1^2+2^2+3^2+\ldots+n^2}{n}$. Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get: $	ext{Mean} = \frac{\frac{n(n+1)(2n+1)}{6}}{n} = \frac{(n+1)(2n+1)}{6}$. Expanding the numerator: $\frac{2n^2+n+2n+1}{6} = \frac{2n^2+3n+1}{6}$.

The mean of the squares of the first $n$ natural numbers is

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