Polynomials and the IMO I have been stuck on this problem, as I am unable to proceed properly. The problem is as follows:  
If $f(x) = (x+2x^2+\cdots nx^n)^2 = a_2x^2 + a_3x^3 +\cdots a_{2n}x^{2n},$ prove that $$a_{n+1} + a_{n+2} +\cdots +a_{2n} = \binom{n+1}{2}\frac{5n^2+5n+2}{12}$$    
I tried expanding the LHS but only ended up with a bunch of expressions. Any help is appreciated.

Hint: Induction works. Let $S_n$ be the sum on the LHS of the desired inequality. Then 
$$
S_1=1=\binom{1+1}{2}\frac{5\cdot 1^2+5\cdot 1+2}{12}.
$$ 
From
\begin{aligned}
&\quad(x+2x^2+\cdots+nx^n+(n+1)x^{n+1})^2\\
&=(x+2x^2+\cdots+nx^n)^2+2(n+1)(x+2x^2+\cdots+nx^n)x^{n+1}+(n+1)^2x^{2(n+1)}\\
\end{aligned}
we have
$$
\boxed{S_{n+1}=S_n+2(n+1)\frac{n(n+1)}{2}+(n+1)^2-\varphi_n}\tag{$*$}
$$
where $\varphi_n$ is the cofficient of $x^{n+1}$ in the expansion of $(x+2x^2+\cdots+nx^n)^2$:
\begin{aligned}
\varphi_n&=1\cdot n+2(n-1)+3(n-2)+\cdots+n\cdot 1\\
&=\sum_{i=1}^ni(n+1-i)\\
&=(n+1)\sum_{i=1}^ni-\sum_{i=1}^ni^2\\
&=\frac{1}{6}n(n+1)(n+2).
\end{aligned}
