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