How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of 
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively. 
 A: For convenience, in the index notation, I only sum over nonnegative integers.
\begin{align}
\sum_{6r_1+3r_2+r_3=24} (1+r_1) &= \sum_{2r_1+r_2+k=8} (r_1+1) \text{, we let } r_3 = 3k \\
&= \sum_{2r_1+2w=8} (2w+1)(r_1+1) \text{, we let } r_2+k = 2w \\
&= \sum_{r_1+w=4} (2w+1)(r_1+1)\\
&=\sum_{w=0}^4(2w+1)(5-w) \\
&= -2 \sum_{w=0}^4w^2 + 9 \sum_{w=0}^4w+5 \sum_{w=0}^4 1 \\
&= -2 \cdot \frac{4(5)(9)}{6} + 9\cdot \frac{4(5)}{2}+ 5 \cdot 5 \\
&= - 60 +90+ 25 \\
&= 55
\end{align}
A: The given rational function can be written as
$$(1+x^6+x^{12}+\ldots)(1+x^6+x^{12}+\ldots)(1+x^3+x^6+\ldots)(1+x^2+x^3+\ldots)\ .$$
The coefficient $c_{24}$ we are after therefore is the number of nonnegative integer solutions to
$$6k_1+6k_2+3k_3+k_4=24\ ,$$
or $6k_1+6k_2+3k_3\leq24$, which is the same as
$$2(k_1+k_2)+k_3\leq8\ .$$
If $k_1+k_2=:j\geq0$ each given value of $j$ ca be realized by $k_1$ and $k_2$ in $j+1$ ways, and then $k_3$ may take $8-2j+1$ different values. It follows that
$$c_{24}=\sum_{j=0}^4(j+1)\cdot(9-2j)=1\cdot9+2\cdot7+3\cdot5+4\cdot 3+5\cdot1=55\ .$$
A: $(1+2x^6+3x^{12}+4x^{18}+5x^{24}+.....).(1+x^3+x^6+x^9+x^{12}+.....x^{24}+.....).(1+x+x^2+.....x^{24}+.....)$
now collect all terms factors which yield $x^{24}$
coff. of $x^{24}$ will be$=1\times(5.1.1)+3\times(4.1.1)+5\times(3.1.1)+7\times(2.1.1)+9\times(1.1.1)=55$ 
think a bit and don't miss any of cases in the problem and you'll always get
right answer without any mechanical approach
