I was not formally taught how to evaluate Riemann sums using the summation rule, and so I am going off of solutions to other problems to apply to my own problem. However, I am stuck. Any help would be greatly appreciated.
~Edited per comments below
Question $$\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \Big( 2 + \frac{6i}{n} \Big)^{-2} = $$ Let $x = \frac{i}{n}$. So, now we need to find our upper and lower bounds, which are given by: $$\frac{b-a}{n} = \frac{1}{n}$$
Therefore, $a =0, b=1$. Now, we can convert our Riemann Sum into an integral:
$$\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \Big( 2 + \frac{6i}{n} \Big)^{-2} = \int_{0}^{1} (2 + 6x)^{-2}dx$$
Let $u = 2 + 6x$, so $du = 6 dx$. Our bounds now change from $u = 2 + 6(0) = 2$ and $u = 2 + 6(1) = 8$.
Now, $$\int_{2}^{8} \frac{1}{6} \cdot \frac{1}{u^{2}} du = \frac{1}{6} \int_{2}^{8} \frac{1}{u^{2}} du = - \frac{1}{6} \cdot \frac{1}{u} |_{2}^{8} $$ $$ = \frac{1}{6} \Big[ \frac{1}{2} - \frac{1}{8} \Big]$$ $$ = \frac{1}{16}.$$
Hint:$$\lim_{n\rightarrow\infty}\frac1n\sum_{k=1}^\infty f\left(\frac kn\right)=\int_0^1f(x)dx$$