## Precalculus (10th Edition)

$-1320$
I know that $\sum_{k=1}^{n} (k)=\frac{n(n+1)}{2}.$ Hence $\sum_{k=1}^{40} (-2k+8)=\sum_{k=1}^{40} (-2k)+\sum_{k=1}^{40} (8)=-2\sum_{k=1}^{40} (k)+40(8)=-2\frac{40(40+1)}{2}+320=-1640+320=-1320$