Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 300: 41

$12$

Here, we have $f(x)=1+x^2$ Then, $\Sigma_{i=1}^n (\dfrac{3}{n}) (1+c_i)^2=(\dfrac{27}{n^3}) \Sigma_{i=1}^n (3+i^2)$ or, $3+(\dfrac{27n(n+1)(2n+1)}{6n^3})-(\dfrac{1}{n^3}) \Sigma_{i=1}^n i^2)=\dfrac{18+27/n+9/n^2}{2}+3$ Thus, $\Sigma_{i=1}^n (\dfrac{3}{n}) (1+c_i)^2=9+3=12$