The sum is ∑ i=1 to n of $\displaystyle (2\frac{2i}{n} +1)*\frac{2}{n}$

multiplying it out:

(*formatting is weird, there should be an n and i=1 around the ∑ and the limit goes all the way down)

lim

n-> infinity ∑ $\displaystyle \frac{8i}{n^2} +\frac{2}{n}$

n-> infinity ∑$\displaystyle \frac{8}{n^2}$∑i + $\displaystyle \frac{2}{n}$

n-> infinity ∑$\displaystyle \frac{8}{n^2}[\frac{n(n+1)}{2}] + \frac{2}{n}$

= $\displaystyle \frac{8}{2}$

$\displaystyle =4$

Answer should be6.