Answer
$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{2 i}{n}\right)^{2}\left(\frac{2}{n}\right) =\frac{26}{3}$$
Work Step by Step
Given $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{2 i}{n}\right)^{2}\left(\frac{2}{n}\right) $$
So, we get
\begin{aligned}
L&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(1+\frac{2 i}{n}\right)^{2}\left(\frac{2}{n}\right) \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{n+2 i}{n}\right)^{2}\left(\frac{2}{n}\right) \\
&
=\lim _{n \rightarrow \infty} \frac{2}{n^{3}} \sum_{i=1}^{n}(n+2 i)^{2} \\
&=\lim _{n \rightarrow \infty} \frac{2}{n^{3}} \sum_{i=1}^{n}(n^2+4 n i+4 i^2) \\
&=\lim _{n \rightarrow \infty} \frac{2}{n^{3}}\left[\sum_{i=1}^{n} n^{2}+4 n \sum_{i=1}^{n} i+4 \sum_{i=1}^{n} i^{2}\right]
\end{aligned}
since $$\sum_{i=1}^{n} n^{2}= n^{3}, \ \ \sum_{i=1}^{n} i=\frac{n(n+1)}{2} , \ \ \\ \sum_{i=1}^{n} i^{2}=\frac{(n)(n+1)(2 n+1)}{6} $$
so, we get
\begin{aligned}
L &=\lim _{n \rightarrow \infty} \frac{2}{n^{3}}\left[n^{3}+(4 n)\left(\frac{n(n+1)}{2}\right)+\frac{4(n)(n+1)(2 n+1)}{6}\right] \\
&=\lim _{n \rightarrow \infty} \frac{2}{n^{3}}\left[n^{3}+2 n^3+2n^2+\frac{4 n^3}{3}+2 n^2+\frac{2 n}{3}\right]\\
&=\lim _{n \rightarrow \infty} \frac{2}{n^{3}}\left[\frac{13 n^3}{3}+4 n^2+\frac{2 n}{3}\right]\\
&=\lim _{n \rightarrow \infty}\left[\frac{4}{3 n^2}+\frac{8}{n}+\frac{26}{3}\right], \ \ \ (since \lim _{n \rightarrow \infty} \frac{1}{ n}=0)\\
&=\frac{26}{3} \end{aligned}
