Don't worry about it.

$\displaystyle =\frac{6}{n}\sum_{i=1}^{n}\left(-125+\frac{225i}{n}-\frac{135i^2}{n^2}+\frac{27i^3}{n^3}\right)$

Now, from here you want to get all of the i's alone. So...

Since the sum of a sum is the sum of the sums, we can kinda pass the sigma throughout

$\displaystyle \frac{6}{n}\left[-\sum_{i=1}^{n}(125)+\sum_{i=1}^{n}\left(\frac{225i }{n}\right)-\sum_{i=1}^{n}\left(\frac{135i^2}{n^2}\right)+\sum _{i=1}^{n}\left(\frac{27i^3}{n^3}\right)\right]$

$\displaystyle \frac{6}{n}\left[-125n+\frac{225}{n}\sum_{i=1}^{n}i-\frac{135}{n^2}\sum_{i=1}^{n}i^2+\frac{27}{n^3}\su m_{i=1}^{n}i^3\right]$

Now, from the formulas for these special sums...