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

The mean of the squares of first $n$ natural numbers is
  1. $\left[\frac{n(n+1)}{2}\right]^2$
  2. $\frac{2 n^2-3 n+1}{6}$
  3. $\frac{2 n^2+3 n+1}{6}$
  4. $\frac{n(n+1)(2 n+1)}{6}$

Solution

First $n$ natural numbers $ 1,2,3,4, \ldots, n $ Their squares $1^2, 2^2, 3^2, \ldots, n^2$ $ \text { Now mean }=\frac{\text { Sum of observations }}{\text { Total number of observations }} $ $ \begin{aligned} & =\frac{1^2+2^2+3^2+\ldots+n^2}{n} \\ & =\frac{\frac{n}{6}(n+1)(2 n+1)}{n}\left[\because \Sigma n^2=\frac{n}{6}(n+1)(2 n+1)\right] \\ & =\frac{(n+1)(2 n+1)}{6}=\frac{2 n^2+3 n+1}{6} \end{aligned} $

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

Practice more Statistics questions on Aicharya