Answer
$\displaystyle \frac{3}{n}\sum_{k=1}^{n}[2(1+\frac{3k}{n})^{2}]$
Work Step by Step
There are n terms in the sum (we don't know n yet)
Each term has the same form:
$[2(1+\displaystyle \frac{3k}{n})^{2}(\frac{3}{n})]$
with the 2(1+... , the denominator n, and $(\displaystyle \frac{3}{n})$ appearing in all the terms unchanged,
the numerators $3k$ assume values
$3,6,9,...,3n$
$3(1),3(2),3(3),...,3\mathrm{n}$
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 $(\displaystyle \frac{3}{n})$ can be factored out of each term of the sum
Using k for indexing,
Sum = $\displaystyle \frac{3}{n}\sum_{k=1}^{n}[2(1+\frac{3k}{n})^{2}]$
