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The mean of the squares of the first n natural numbers is
$$\frac{1}{6}(2n^2 + 3n + 1)$$
$$\frac{1}{3}(2n^2 + 3n + 1)$$
$$\frac{1}{6n}(2n^2 + 3n + 1)$$
$$\frac{1}{6} n (n + 1)(2n + 1)$$
The mean of first n natural numbers = [n(n+1)]/2
The mean of square of first n natural nos.=[(n)(n+1)(2n+1)]/6
= [2n^2 + 3n + 1]/6
The mean of cube first n natural numbers = {[n(n+1)]/2}^2
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