I am trying to solve one of the exercises my professor told us to try, just to understand how Integral calculations work using Riemann Sums instead of antiderivatives.
However, I got stuck as I do not know how to continue simplifying...
$\lim_{n \to \infty} \sum_{i=1}^n [(-1+ \frac{4i}{n})^3 + 1]\frac{4}{n}$
$\lim_{n \to \infty} \frac{4}{n}\sum_{i=1}^n [(-1+ \frac{4i}{n})^3 + 1]$
$\lim_{n \to \infty} \frac{4}{n}[\sum_{i=1}^n (-1+ \frac{4i}{n})^3 + \sum_{i=1}^n 1]$
$\lim_{n \to \infty} \frac{4}{n}[\sum_{i=1}^n (-1+ \frac{4i}{n})^3 + n]$
$\lim_{n \to \infty} \frac{4}{n}[(-1+ \frac{4}{n})^3\sum_{i=1}^n (i^3) + n]$
$\lim_{n \to \infty} \frac{4}{n}[(-1+ \frac{4}{n})^3( \frac{n^2(n+1)}{4}) + n]$
$\lim_{n \to \infty} [(\frac{-4}{n}+ \frac{16}{n^3})^3( \frac{n^2(n+1)}{4}) + 4]$
$\lim_{n \to \infty} [(\frac{-4}{n}+ \frac{16}{n^3})^3(\frac{n^4+2n^3+2n^2}{4}) + 4]$
This is where I got stuck. I have no idea what to do with the first parenthesis, especially because of the exponent 3...
Thank you for your help in advance. I am also ready to give explanation on my reasoning or calculations if needed.
HINT:
$$\left(-1+\frac{4i}{n} \right)^3=-1+\frac{12i}{n} -\frac{48i^2}{n^2}+\frac{64i^3}{n^3}$$