Answer
$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(2+\frac{3 i}{n}\right)^{3}\left(\frac{3}{n}\right) =\frac{609}{4}$$
Work Step by Step
Given$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(2+\frac{3 i}{n}\right)^{3}\left(\frac{3}{n}\right) $$,
so, we have
\begin{aligned}
L&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(2+\frac{3 i}{n}\right)^{3}\left(\frac{3}{n}\right) \\
&=\lim _{n \rightarrow \infty} \frac{3}{n} \sum_{n=1}^{n}\left[\frac{2 n+3 i}{n}\right]^{3} \\
&=\lim _{n \rightarrow \infty} \frac{3}{n^4} \sum_{n=1}^{n}\left[2 n+3 i\right]^{3} \\
&=\lim _{n \rightarrow \infty} \frac{3}{n^{4}} \sum_{i=1}^{n}\left(8 n^{3}+36 n^{2} i+54 n i^{2}+27 i^{3}\right) \\
\end{aligned}
$$Since \sum_{i=1}^{n} 1=n, \ \sum_{i=1}^{n} i=\frac{n(n+1)}{2},\\ \ \ \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}, \ \ \sum_{i=1}^{n} i^{3}= \frac{n^{2}(n+1)^{2}}{4}$$
so, we get
\begin{aligned}
L &=\lim _{n \rightarrow \infty} \frac{3}{n^{4}}\left(8 n^{4}+36 n^{2} \frac{n(n+1)}{2}+54 n \frac{n(n+1)(2 n+1)}{6}+27 \frac{n^{2}(n+1)^{2}}{4}\right)
\\
&=\lim _{n \rightarrow \infty} 3\left(8+18 \frac{(n+1)}{n}+\frac{9(n+1)(2 n+1)}{n^{2}}+\frac{27}{4} \cdot \frac{(n+1)^{2}}{n^{2}}\right) \\
&=\lim _{n \rightarrow \infty} 3\left(8+18 \frac{(n+1)}{n}+\frac{9(2n^2+3n+1)}{n^{2}}+\frac{27}{4} \cdot \frac{n^{2}+2n+1}{n^{2}}\right) \\
&=\lim _{n \rightarrow \infty} 3\left(8+18(1+ \frac{1}{n})+18+\frac{27}{n}+\frac{9}{n^{2}}+\frac{27}{4}(1+ \frac{2}{n}+\frac{1}{n^{2}})\right) \\
&=3\left(8+18+18+\frac{27}{4}\right), \ (as \lim _{n \rightarrow \infty}\frac{1}{n}=\lim _{n \rightarrow \infty}\frac{1}{n^2}=0)\\
&=\frac{609}{4}
\end{aligned}
