It's so big!

I treated this problem as an Arithmetic Progression. It has 2017 terms. The first term is 3 and the common difference is 3.

$S=\frac{n}{2}[2a_1 + (n-1)]d=\frac{2017}{2}[2(3) + (2016)(3)]=6105459$

The sum can be written as $\displaystyle{\sum_{n=1}^{2017}{3n}} = \displaystyle{3\sum_{n=1}^{2017}{n}}$ . This simplifies to $\displaystyle{3(1+2+3+\cdots+2017) \\ \implies 3\dfrac{(2017)(2018)}{2} \\ = \boxed{6105459.}}$