<s>[INST] Help trying to find the coefficient in a generating function expansion Find the coefficient of ${x^{20}}$ in the expansion of the generating function $g(x) = \frac{5{(1-x^5)^7}}{(1-x)^{2}}$
I broke the function into two components: $5{(1-x^5)^7}$  and $\frac{1}{(1-x)^2}$
Because I'm looking for the ${x^{20}}$ coefficient, I have 5 terms:
$(a_0 \cdot b_{20}) + (a_5 \cdot b_{15}) + (a_{10} \cdot b_{10}) + (a_{15} \cdot b_5) + (a_{20} \cdot b_0)$
which gives me:
$20+2-1 \choose 20$ $-$ $7 \choose 1$ $15+2-1 \choose 15$$ $+$ $$7 \choose 2$$10+2-1 \choose 10$$ $-$ $$7 \choose 3$$45+2-1 \choose 5$$ $+$ $$7 \choose 4$$0+2-1 \choose 0 $
I believe that I multiply the $5$ in the first polynomial to the coefficient I find, but my answer comes out to be $(5 \cdot (-35))=-175$ which I don't believe is possible. Is this the correct answer?
 [/INST] Yes you are correct. We have that 
$$(1-x^5)^7=1-7x^5+21x^{10}-35x^{15}+35x^{20}+o(x^{20}).$$
and
$$\frac{1}{(1-x)^{2}}=\sum_{n=0}^{\infty} (n+1)x^n.$$
Therefore 
$$[x^{20}]\frac{(1-x^5)^7}{(1-x^2)^{2}}=1\cdot (20+1)-7\cdot (15+1)+21\cdot (10+1)-35\cdot (5+1)+35\cdot 1=-35$$
So the desired coefficient is $5\cdot (-35)=-175$.
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