Question

Find mean of Squares of the first n natural numbers

Answer 1

Hi,
$$\begin{gathered} \mathrm{Sum}\mathrm{of}\mathrm{square}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{natural}\mathrm{number} \\ =1^2+2^2+3^2+.....+{\mathrm{n}}^2 \\ =\frac{1}{6}\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right) \\ \mathrm{So}\mathrm{mean}=\frac{\mathrm{Sum}}{\mathrm{n}}=\frac{1}{6\mathrm{n}}\times \mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)=\frac{1}{6}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right) \end{gathered}$$
Regards