Solving the Riemann Sum $\sum_{i=1}^{n}(1+\frac{6i}{n})^3(\frac{2}{n})$?

Question

So I have the Riemann sum. $\sum_{i=1}^{n}(1+\frac{6i}{n})^3(\frac{2}{n})$. From my understanding that turns into $(\frac{2}{n})\sum_{i=1}^{n}(1+\frac{6i}{n})^3$ and what is really perplexing me is that that summation turns into $$ \left(\frac{2}{n}\right)\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^3\rightarrow \left(\frac{2}{n^4}\right)\sum_{i=1}^{n}(n+6i)^3$$From there on out I can solve the summation, but I'm having trouble understanding how the $n$ was pulled out and the new summation was formed. If someone could explain that bit I would greatly appreciate it, thanks in advance!

Answer 1

$$\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{n^{3}}{n^{3}}\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{2}{n\cdot n^{3}}\sum_{i=1}^{n}n^{3}\left(1+\frac{6i}{n}\right)^{3}$$ and $$n^{3}\left(1+\frac{6i}{n}\right)^{3}=\left(n+\frac{6in}{n}\right)^{3}=\left(n+6i\right)^{3}$$

Answer 2

$(\frac{2}{n})\sum_{i=1}^{n}(1+\frac{6i}{n})^3=(\frac{2}{n})\sum_{i=1}^{n}(\frac1{n^3})(n+6i)^3=(\frac{2}{n^4})\sum_{i=1}^{n}(n+6i)^3$. You can pull out the $\frac1{n^4}$ for the same reason you can pull out the $\frac2n$; because that value is constant for all $i$, it's the same as just multiplying every part of the sum by it and factoring it out.

Answer 3

$(\frac{2}{n})\sum_{i=1}^{n}(1+\frac{6i}{n})^3=(\frac{2}{n})\sum_{i=1}^{n}(\frac{n+6i}{n})^3=(\frac{2}{n^4})\sum_{i=1}^{n}(n+6i)^3$$