Answer
$\displaystyle \frac{2}{n}\sum_{k=1}^{n}\left[\left(\frac{2k}{n}\right)^{3}-\left(\frac{2k}{n}\right)\right]$
Work Step by Step
There are n terms in the sum (we don't know n yet)
Each term has the same form:
$[(\displaystyle \frac{2k}{n})^{3}-(\frac{2k}{n})(\frac{2}{n})]$
with n and the last numerator,2, appearing in all the terms unchanged,
The numerators 2k inside the brackets assume values
$2,4,6,...2n$
$2(1),2(2),2(3),...,2\mathrm{n}$
So, each term varies as k varies from 1,2,3,...n.
We still don't know n, but once it is known,
it is constant throughout the sum,
so the factor $(\displaystyle \frac{2}{n})$ can be factored out of each term of the sum
Using k for indexing,
Sum = $\displaystyle \frac{2}{n}\sum_{k=1}^{n}\left[\left(\frac{2k}{n}\right)^{3}-\left(\frac{2k}{n}\right)\right]$
