Answer
$308$
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 $\Sigma_{k=1}^{7} k(2k+1)=\Sigma_{k=1}^{7} 2k^2+\Sigma_{k=1}^{7} k$
or, $2[\dfrac{7 \cdot 8 \cdot 15}{6}]+\dfrac{7 \cdot 8}{2}=280+28=308$