Answer
$588$
Work Step by Step
Use formula: $\Sigma_{k=1}^n k=\dfrac{n(n+1)}{2}$ and $\Sigma_{k=1}^n k^2=\dfrac{n(n+1)(n+2)}{6}$
Here, we have $(\Sigma_{k=1}^{7} k)^2+\Sigma_{k=1}^{7} \dfrac{k^3}{4}=(\dfrac{7 \cdot 8}{2})^2-\dfrac{1}{4}\Sigma_{k=1}^{7} k^3$
or, $(28)^2-(\dfrac{1}{4})(28)^2=588$
