## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-1320$
We know that $\sum_{k=1}^{n} \ k =\dfrac{n(n+1)}{2}$ Plug in $40$ for $n$ in the given summation formula. Therefore, $\sum_{k=1}^{40} (-2k+8) =\sum_{k=1}^{40} (-2k) +\sum_{k=1}^{40} \ 8 \\ =-2 \sum_{k=1}^{40} k +(40) \ (8) \\=(-2) \times \dfrac{40(40+1)}{2}+320\\=-1640+320 \\=-1320$