The sum, ∑ n = 1 7 n n + 1 2 n + 1 4 , is equal to

Given,

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

$=\frac{1}{4}\left[\sum _{n=1}^{7}2n^3+3n^2+n\right]$

Now using $\sum _{}^{}\left(1+2+3+...+n\right)=\frac{n\left(n+1\right)}{2}$, $\sum _{}^{}\left(1^2+2^2+3^2+...+n^2\right)=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ and $\sum _{}^{}\left(1^3+2^3+3^3+...+n^3\right)={\left[\frac{n\left(n+1\right)}{2}\right]}^2$, we can write
$=\frac{1}{4}\left[2{\left(\frac{7·8}{2}\right)}^2+3\left(\frac{7·8·15}{6}\right)+\frac{7·8}{2}\right]$
$\frac{1}{4}\left[2\times 49\times 16+28\times 15+28\right]$
$\frac{1}{4}\left[1568+420+28\right]==504$.