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!

Let $S_n$ be the required sum. By expanding the right hand side of $$a_0 + a_1 x + \ldots + a_{2n} x^{2n} = (x + 2x^2 + \ldots + nx^n)^2$$ we have 
$$a_{n+i} = n \cdot i + (n-1) \cdot (i+1) + (n-2) \cdot (i+2) + \cdots \ + i \cdot n$$ for $i=1, 2, \ldots, n.$ 
Summing over $i=1,2,\ldots, n$ gives
$$ S_n = n(1 + 2 + 3+ \ldots + n) + (n-1)(2+ 3 + \ldots + n) + \ \ldots \ + 2( (n-1) + n) + 1(n) $$
The terms in the brackets are the sum of the first $n$ integers (which equals $\binom{n+1}{2}$) minus the sum of the first $k$ integers (which equals $\binom{k+1}{2}$) so we get
$$ S_n = \sum_{k=0}^{n-1} (n-k) \left( \binom{n+1}{2} - \binom{k+1}{2} \right)$$
$$ = \binom{n+1}{2} \sum_{k=0}^{n-1} (n-k) - \sum_{k=1}^{n-1} \binom{n-k}{1} \binom{k+1}{2}$$
Again, we have $\sum_{k=0}^{n-1} (n-k) = \binom{n+1}{2}$ so the first term above simplifies to $\binom{n+1}{2}^2.$ 
To pick a $4$-element subset from $n+2$ elements we follow this scheme - Pick element $k+2$  ($ \ k$ from $1$ to $n-1$) to be $3$-rd element of the subset, then pick $2$ from the $k+1$ elements on its left and $1$ element from the $n-k$ on its right. Summing over all possibilities of what the $3$rd element could be yields the second term above.
Therefore, $$S_n = \binom{n+1}{2}^2 - \binom{n+2}{4}$$
$$ \ \ \ \ \ \  \ \ \  = \frac {n(n+1)(5n^2+5n+2)} {24}$$
