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. 

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}
