Sum of coefficients of $x^i$ (Multinomial theorem application) 
A polynomial in $x$ is defined by
  $$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
  Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
  $$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$

I don't know how to proceed. I know the Multinomial theorem, however, I have problems in applying it. Any help will be appreciated as it will help me understand the theorem well.
Thanks!

Write $(x+2x^2+\cdots+nx^n)^2=(x+2x^2+\cdots+nx^n)(x+2x^2+\cdots+nx^n)$ and for each coefficient in the first factor find the sum of the coefficients in the second factor with which it will enter into the desired sum: The coefficient $k$ in the first factor is paired with $n+1-k$ through $n$ in the second factor, so it contributes
\begin{eqnarray*}
k\sum_{j=n+1-k}^nj
&=&
k\left(\sum_{j=1}^nj-\sum_{j=1}^{n-k}j\right)
\\
&=&
k\left(\frac{n(n+1)}2-\frac{(n-k)(n-k+1)}2\right)
\\&=&
\left(n+\frac12\right)k^2-\frac12k^3\;.
\end{eqnarray*}
Then summing over $k$ yields
\begin{eqnarray*}
\sum_{k=1}^n\left(\left(n+\frac12\right)k^2-\frac12k^3\right)
&=&\left(n+\frac12\right)\sum_{k=1}^nk^2-\frac12\sum_{k=1}^nk^3
\\
&=&
\left(n+\frac12\right)\frac{n(n+1)(2n+1)}6-\frac12\left(\frac{n(n+1)}2\right)^2
\\
&=&
\frac{n(n+1)(5n^2+5n+2)}{24}\;.
\end{eqnarray*}
