Answer
12
Work Step by Step
$$\lim_{n\to\infty}\sum_{i=1}^n\frac{24i}{n^2}$$
$$\sum_{i=1}^n\frac{24i}{n^2}=\frac{24}{n^2}\sum_{i=1}^ni=\frac{24}{n^2}\frac{n(n+1)}{2}$$
$$=\frac{12n(n+1)}{n^2}=\frac{12(n+1)}{n}$$
Therefore,
$$\lim_{n\to\infty}\sum_{i=1}^n\frac{24i}{n^2}=\lim_{n\to\infty}\frac{12(n+1)}{n}=12$$
