Verify the following equation. And, note that index runs over only upto a finite stage. Because, this is false for many infinite cases (as I indicate at the end!)
$\sum_{i=1}^n (a_i+b_i)=\sum_{i=1}^n a_i+\sum_{i=1}^n b_i\\\ \sum_{i=1}^nc\cdot a_i=c\cdot\left(\sum_{i=1}^na_i\right)$
where $c$ is a constant independent of the running index.
So, the desired sum becomes, $\sum_{i=1}^k (i^2+3i)=\sum_{i=1}^ki^2+3\sum_{i=1}^k i$
As you have noted, you can now use, $\displaystyle\sum_{i=1}^ki=\dfrac{k(k+1)}{2}$ and $\displaystyle\sum_{i=1}^ki^2=\dfrac{k(k+1)(2k+1)}{6}$ to get the sum.
Here is what I mean when I talk about the infinite series:
$\sum_{i=1}^\infty (i-i)=\sum_{i=1}^\infty 0=0$ but when you expand the sum as I wrote down above, you'll get $\infty-\infty$ which clearly refuses to make sense!
That you call this complicated, I am inclined to give you some references, which if you're Mathematically inclined, you'll love.
Suggested Reading:
Terry Tao -- Analysis, Volume I, TRIM series, Hindustan Book Agency.(here) for errata and some links
Knopp -- Infinite sequence and Series, Dover Books on Mathematics (here)
Thomas Bromwich -- An Introduction to the theory of Infinite Series (archive.com link)
You may like this answer of mine in connection with the last reference.