Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$1560$
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 rewrite the given sequence and apply the above formulas: $\displaystyle \sum_{k=1}^{16} (k^2+4) = \sum_{k=1}^{16} (k^2) + \sum_{k=1}^{16} (4)=\dfrac{16(16+1) [(2)(16)+1)]}{6}-(16)(4) \\=1496+63 \\= 1560$