Question 141

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

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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