Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 300: 39

$\dfrac{2}{3}$

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