$\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$
I would assume you haven't done the fundamental theorem of calculus yet, because this Riemann sum appears to be $\displaystyle \int_1^4x^3-2x\,dx=\frac{195}{4}$
Pull the $\displaystyle \frac{3}{n}$ out and expand the summand. You get
$\displaystyle \lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left[(3i/n)^3+3(3i/n)^2+(3i/n)-1\right]$ $\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[\frac{27}{n^3}\sum_{i=1}^n i^3+\frac{27}{n^2}\sum_{i=1}^n i^2+\frac{3}{n}\sum_{i=1}^n i-\sum_{i=1}^n 1\right]$
Evaluate that using summation formulas and then take the limit.
I have studied the fundamental thereom although it was quite some time ago, would you mind explaining how you obtained 4 on the upper limit of integration for $\displaystyle \int_1^4x^3-2xdx?$
Sorry, think I may have just found it:
$\displaystyle \frac{3}{n} = \frac{b-1}{n} $ is that the formula?