Finding coefficient of generating function 
Find the coefficient of $x^{52}$ in $$(x^{10} + x^{11} + \ldots + x^{25})(x + x^2 + \ldots + x^{15})(x^{20} + x^{21}+ \ldots + x^{45})$$

One thing I tried doing was factoring out $x^{10}, x,  x^{20}$ from each of the products, respectively, then using the identity of a product of two polynomials to obtain the coefficients...But I ended up with $1$. I can't figure this out and any help would be much appreciated!
 A: The original problem, after factoring out the terms as stated in the post, we have
$$
x^{10}(1+x+\ldots+x^{15})\cdot x(1 + x + \ldots+x^{14})\cdot x^{20}(1+x+\ldots+x^{25}) \tag{1}
$$
Applying some identities, $(1)$ gives
$$ x^{31}\cdot\frac{1-x^{16}}{1-x}\cdot\frac{1-x^{15}}{1-x}\cdot\frac{1-x^{26}}{1-x} \tag{2}$$
Rearranging the terms, we then have
$$
x^{31}\cdot(1-x^{16})\cdot(1-x^{15})\cdot(1-x^{26})\cdot \frac{1}{(1-x)^3} \tag{*}
$$
The last term in the above product is equal to
$$1 + \binom{3}{1}x + \binom{4}{2}x^2 + \ldots \tag3$$
Finally we take the products that can result in $x^{52}$ and obtain the coefficients:

*

*$x^{31}$ (first term in $(*)$) and $x^{21}$ (from $(3)$),

*$x^{31}$, $x^{16}$ (second term in $(*)$), $x^5$ in $(3)$

*$x^{31}$(first term in $(*)$), $x^{15}$ (third term in $(3)$), and $x^6$ in $(3)$
Since all the coefficients of the polynomials equal $1$ or $-1$ except for the polynomial expanded in $(3)$, we have as our coefficient
$$
\binom{21+3-1}{21} - \binom{6+3-1}{6} - \binom{5+3-1}{5} = 204
$$
Note: I hadn't seen Andre's solution prior to typing this.
A: Factoring out is a good beginning. Now use the fact that $1+x+x^2+\cdots+x^{n-1}=\frac{1-x^n}{1-x}$ (if $x\ne 1$).
So your expression will be a power of $x$, times a product of three polynomials of shape $1-x^k$, divided by $(1-x)^3$.
Use the generalized Binomial Theorem (or Taylor series) to find the power series for $\dfrac{1}{(1-x)^3}$. Putting the pieces together is slightly unpleasant, but doable. 
