Find the closed form of a summation from $k=1$ to $n$ For my discrete mathematics class, I need to express this summation in closed form in terms of $n$,
$$\sum_{k=1}^n \left(6 + 2 \cdot \frac{k}{n}\right)^2. $$
I was in the hospital when they went over this in class, and they were not working out of the textbook. I've looked extensively online and found plenty of cases where people solved problems like this, but no one explained how they did it. 
I know the first thing to do is to multiply out the expression, but I'm not sure how to proceed from there. Any help would be hugely appreciated. 

Based on the responses you all have given me, I have an idea of how I should solve this. Can anyone verify that this is correct?
so:
$$\sum_{k=1}^n \left(6 + 2 \cdot \frac{k}{n}\right)^2 $$
$$\equiv \sum_{k=1}^n 36 + \sum_{k=1}^n \frac{24k}{n} + \sum_{k=1}^n \frac{4k^2}{n^2}$$
$$\equiv 36 + \frac{24}{n} \sum_{k=1}^n k + \frac{4}{n^2} \sum_{k=1}^n k^2$$
$$\equiv 36 + \frac{24}{n} \cdot \frac{n(n + 1)}{2} + \frac{4}{n^2} \cdot\frac{n(n + 1)(2n + 1)}{6}$$
..and then simplify? Is that correct? Thanks!

Just do it.
$(6 + 2\frac kn)^2 = 36 + \frac 24n k + \frac 4{n^2}k^2$
So $\sum_{k=1}^n (6 + 2\frac kn)^2=\sum_{k=1}^n(36 + \frac 24n k + \frac 4{n^2}k^2)=$
$\sum_{k=1}^n 36 + \frac 24n\sum_{k=1}^n k + \frac4{n^2}\sum_{k=1}^n k^2=$
$36n + \frac 24n\frac {n(n+1)}2 +  \frac 4{n^2}\frac{n(n+1)(2n+1)}6=$
$36n +12(n+1) +\frac 2{3n}(n+1)(2n+1)=$
$\frac {144n^2 + 12n +2(n^2+ 3n+1)}{3n}=$
$\frac {146n^2 + 18n + 2}{3n}$
