Question

The sum $\sum _{n=1}^7\frac{n\left(n+1\right)\left(2n+1\right)}{4}$ is equal to

Answer 1

Evaluate the given expression

Given, $\sum _{n=1}^7\frac{n\left(n+1\right)\left(2n+1\right)}{4}$

$\begin{array}{cll}& \sum _{n=1}^7\frac{n\left(n+1\right)\left(2n+1\right)}{4}& \\ =& \frac{1}{4}\sum _{n=1}^{7}n\left(n+1\right)\left(2n+1\right)& ..\mathbf{.}\mathbf{[}\mathbf{\because }\sum _i\mathbf{k}\mathbf{i}\mathbf{=}\mathbf{k}\sum _i\mathbf{i}\mathbf{]}\\ =& \frac{1}{4}\sum _{n=1}^7\left(2n^3+3n^2+n\right)& \\ =& \frac{1}{4}\left[\sum _{n=1}^{7}2n^3+\sum _{n=1}^{7}3n^2+\sum _{n=1}^{7}n\right]& \\ =& \frac{1}{4}\left[2\sum _{n=1}^7{n}^3+3\sum _{n=1}^7{n}^2+\sum _{n=1}^{7}n\right]& \\ =& \frac{1}{4}\left[2{\overline{){\left[\frac{n\left(n+1\right)}{2}\right]}^2}}_1^7+3{\overline{)\frac{n\left(n+1\right)\left(2n+1\right)}{6}}}_1^7+{\overline{)\frac{n\left(n+1\right)}{2}}}_1^7\right]& ...\mathbf{[}\mathbf{\because }\mathbf{\sum }{\mathbf{k}}^{\mathbf{3}}\mathbf{=}{\mathbf{\left[}\frac{\mathbf{k}\left(k+1\right)}{\mathbf{2}}\mathbf{\right]}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{\sum }{\mathbf{k}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{k}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{,}\mathbf{}\mathbf{\sum }\mathbf{k}\mathbf{=}\frac{\mathbf{k}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{]}\\ =& \frac{1}{4}\left[2{\left(\frac{7\times 8}{2}\right)}^2+3\left(\frac{7\times 8\times 15}{6}\right)+\frac{7\times 8}{2}\right]& \\ =& 504& \end{array}$

Therefore, $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{7}}\frac{\mathbf{n}\left(n+1\right)\left(2n+1\right)}{\mathbf{4}}\mathbf{=}\mathbf{504}$.