<s>[INST] How to rewrite a Sum? How to find the following sum?
$$
\frac{2}{n}  \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right)
$$
For example
$$
\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.
$$
I need to find the value of which this would converge to and I know that it must be six because the integral from $1$ to $3$ of the function $(2x-1)$ is $6$.
Any help is appreciated!!!
 [/INST] $$\frac{2}{n}  \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) = \frac{2}{n}\left( \sum_{i=1}^{n} \frac{4i}{n} + \sum_{i=1}^{n} 1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i + \frac{2}{n}\cdot n =  \frac{8}{n^2}\cdot \frac{n(n+1)}{2} + 2 = 4 \left(1+\frac{1}{n} \right) + 2$$
</s>