Answer
$-2376$
Work Step by Step
... we want to apply (1) $\displaystyle \sum_{k=1}^{n}(ca_{k})=c\sum_{k=1}^{n}a_{k}$, but the index does not start at 1.
$\displaystyle \sum_{k=8}^{40}(-3k)=$ (terms from 8 to 40) = (40 terms) - (first 7 terms)
$=\displaystyle \sum_{k=1}^{40}(-3k)-\sum_{k=1}^{7}(-3k)$
... apply formula (1)
$=(-3)\displaystyle \sum_{k=1}^{40}k-(-3)\sum_{k=1}^{7}k$
... apply $6.\displaystyle \qquad \sum_{k=1}^{n}k=1+2+3+\cdots+n=\frac{n(n+1)}{2}$
$=-3\displaystyle \left[\frac{40(40+1)}{2}-\frac{7(7+1)}{2}\right]$
$=-3[820-28]$
$=-2376$