## College Algebra (10th Edition)

$3570$
... 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. $\displaystyle \sum_{k=10}^{60}2k=$ (terms from 10 to 60) = (60 terms) - (first 9 terms) $=\displaystyle \sum_{k=1}^{60}2k-\sum_{k=1}^{9}2k$ ... apply formula (1) $=2\displaystyle \sum_{k=1}^{60}k-2\sum_{k=1}^{9}k$ ... apply $6.\displaystyle \qquad \sum_{k=1}^{n}k=1+2+3+\cdots+n=\frac{n(n+1)}{2}$ $=2\displaystyle \left[\frac{60(60+1)}{2}-\frac{9(9+1)}{2}\right]$ $=2[1830-45]$ $=3570$