Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 300: 42

$1$

Here, we have $f(x)=3x^2$ Then, $\Sigma_{i=1}^n (\dfrac{3}{n}) (c_i)^2=(\dfrac{3}{n^3}) \Sigma_{i=1}^n (i^2)$ or, $(\dfrac{3}{n^3})(\dfrac{n(n+1)(2n+1)}{6})=\dfrac{2n^3+3n^2+n}{2n^3}$ Thus, $\Sigma_{i=1}^n (\dfrac{3}{n}) (c_i)^2=\lim\limits_{n \to \infty}\dfrac{2+3/n+1/n^2}{2}=1$