Answer
$\dfrac{5}{6}$
Work Step by Step
Consider $f(x)=x+x^2$
This implies that
$\Sigma_{i=1}^n (\dfrac{1}{n}) (k_i+k_i^2)=(\dfrac{1}{n}) \Sigma_{i=1}^n (\dfrac{i}{n}+\dfrac{i^2}{n^2})$
$(\dfrac{1}{n^3})\Sigma_{i=1}^n i+(\dfrac{1}{n^3})\Sigma_{i=1}^n i^2=\dfrac{5n^3+6n^2+n}{6n^3}=\dfrac{5+\dfrac{6}{n}+\dfrac{1}{n^2}}{6}$
Now, $\lim\limits_{n \to \infty}\dfrac{5+\dfrac{6}{n}+\dfrac{1}{n^2}}{6}=\dfrac{5}{6}$
