Answer
$$\frac{9}{2}$$
Work Step by Step
$$\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{3i}{n}\right)\left(\frac{3}{n}\right)$$
$$\sum_{i=1}^n\left(\frac{3i}{n}\right)\left(\frac{3}{n}\right)=\sum_{i=1}^n\left(\frac{9i}{n^2}\right)=\frac{9}{n^2}\sum_{i=1}^ni$$
$$=\frac{9}{n^2}\times\frac{n(n+1)}{2}=\frac{9(n+1)}{2n}$$
$$\lim_{n\to\infty}\frac{9(n+1)}{2n}=\frac{9}{2}$$
