Answer
$$
3
$$
Work Step by Step
We will write the sum in closed form
$\sum_{k=1}^{n} \frac{1+6 k}{n^{2}}=\frac{6}{n^{2}} \sum_{k=1}^{n} \frac{1}{n^{2}} +k\sum_{k=1}^{n} 1$
$=\frac{6}{n^{2}} \cdot \frac{(1+n)n}{2}+n \cdot \frac{1}{n^{2}} $
$=\frac{3 n+1 n+3 n^{2}}{n^{2}}=\frac{4n +3 n^{2}}{n^{2}}$
Take the limit as $n \rightarrow+\infty$
$\lim _{n \rightarrow+\infty} \frac{4 n+3 n^{2}}{n^{2}}=3$
