Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 299: 26

$308$

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, $2[\dfrac{7 \cdot 8 \cdot 15}{6}]+\dfrac{7 \cdot 8}{2}=280+28=308$