Q75 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Mathematics

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

  • A
    $\left[\frac{n(n+1)}{2}\right]^2$
  • B
    $\frac{2 n^2-3 n+1}{6}$
  • Correct answer
    C
    $\frac{2 n^2+3 n+1}{6}$
  • D
    $\frac{n(n+1)(2 n+1)}{6}$

Solution

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

Hence, the correct answer is option: 'C'.

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