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

$-2376$
Recall the formulas: $\displaystyle \sum_{k=1}^{n} (ca_k)=c \ \sum_{k=1}^{n} a_k$ and $\displaystyle \sum_{k=1}^{n} k=\dfrac{n(n+1)}{2}$ We can see that in the given sequence, the index does not start at 1. So, we will re-write the given sequence as: $\sum_{k=8}^{40} (-3k)=$ (Terms from 8 to 40) = (Terms from 1 to 40) - (Terms from 1 to 7) We rewrite the sequence as stated above and apply the sum formulas: $\displaystyle \sum_{k=8}^{40} (-3k)= \sum_{k=1}^{40} (-3k) -\sum_{k=1}^{7} (-3k) \\= (-3) \sum_{k=1}^{40} k - (-3) \sum_{k=1}^{7} k \\= -3 [ \displaystyle \dfrac{40(40+1)}{2} -\dfrac{7(7+1)}{2}] \\=-3 (820 -28) \\= -2376$