If ${\displaystyle \sum _{k=1}^n\left({\displaystyle \sum _{m=1}^k{m}^2}\right)}=a{n}^4+b{n}^3+c{n}^2+dn+e,$  then

$\sum _{k=1}^n\left(\sum _{m=1}^k{m}^2\right)=\sum _{k=1}^n\frac{k(k+1)(2k+1)}{6}$ 

                      $=\frac{1}{6}{\displaystyle \sum _{k=1}^n\left(2k^3+3k^2+k\right)}$

                      =$\frac{1}{3}.{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2+\frac{1}{2}\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{1}{6}\frac{n\left(n+1\right)}{2}$

$a=$ coefficient of $n^4=\frac{1}{3}.\frac{1}{4}$