smc

Let $x=e^{i\pi /6}$, a $30^{\circ }$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is: $$1+2x+3x^2+\cdots +(k+1)x^k$$ We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series. $$\begin{array}{rl}S& =1+2x+3x^2+\cdots +2015x^{2014}\\ xS& =x+2x^2+3x^3+\cdots +2015x^{2015}\\ (1-x)S& =1+x+x^2+\cdots +x^{2014}-2015x^{2015}\\ S& =\frac{1-x^{2015}}{(1-x{)}^2}-\frac{2015x^{2015}}{1-x}\end{array}$$ We want to find $|S|$. First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$. Therefore $$S=\frac{1-\frac{1}{x}}{(1-x{)}^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.$$ Hence, since $|x|=1$, we have $|S|=\frac{2016}{|1-x|}.$ Now we just have to find $|1-x|$. This can just be computed directly: $$1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i$$ $$|1-x{|}^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2$$ $$|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}$$ Therefore $|S|=2016\cdot \frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008\sqrt{6}$. Thus the answer is $1008+2+1008+6=\boxed{2024}$.