Answer
$1560$
Work Step by Step
$\displaystyle \sum_{k=1}^{16}\left(k^{2}+4\right)=$
... use $2.\displaystyle \qquad \sum_{k=1}^{n}(a_{k}+b_{k})=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}$
$=\displaystyle \sum_{k=1}^{16}k^{2}+\sum_{k=1}^{16}4$
... 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 $5.\displaystyle \qquad \sum_{k=1}^{n}c=c+c+\cdots+c=cn$
$=\displaystyle \frac{16(16+1)(2\cdot 16+1)}{6}+4(16)$
$=1496+64$
$=1560$
