Answer
$955$
Work Step by Step
... separate the term (k=0) from the rest
$\displaystyle \sum_{k=0}^{14}(k^{2}-4) =(0^{2}-4)+\displaystyle \sum_{k=1}^{14}(k^{2}-4)$
... use $2.\displaystyle \qquad \sum_{k=1}^{n}(a_{k}+b_{k})=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}$
$=-4+\displaystyle \sum_{k=1}^{14}k^{2}-\sum_{k=1}^{14}4$
... use $7.\displaystyle \qquad \sum_{k=1}^{n}k^{2}=1^{2}+2^{2}+3^{2}+\cdots+n=\frac{n(n+1)(2n+1)}{6}$
... and $5.\displaystyle \qquad \sum_{k=1}^{n}c=c+c+\cdots+c=cn$
$=-4+\displaystyle \frac{14(14+1)(2\cdot 14+1)}{6}-4(14)$
$=-4+1015-64$
$=955$
