Answer
$955$
Work Step by Step
Recall the formulas:
$\displaystyle \sum_{k=1}^{n} k^2=1^1+2^2+...+n=\dfrac{n(n+1) (2n+1)}{6}$
and
$ \sum_{k=1}^{n} c=c+c+c+...+c=cn$
We re-write the given sequence and apply the above formulas:
$\displaystyle \sum_{k=0}^{14} (k^2-4) =-4 + \sum_{k=1}^{14} (k^2-4) \\=-4+ \sum_{k=1}^{14} k^2 -\sum_{k=1}^{14} 4=-4+\dfrac{14(14+1) [(2)(14)+1)]}{6}-(4)(14) \\=-4+1015-64 \\= 955$
