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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:

$(a_0*b_{20})$ + $(a_5*b_{15})$ + $(a_{10}*b_{10})$ + $(a_{15}*b_5$) + $(a_{20}*b_0)$

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 $(5 * -35)=-175$ which I don't believe is possible. Is this the correct answer?

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  • $\begingroup$ Your idea is right, there is just a small typo though it should be ${5+2-1\choose 5} $. BTW how did you get that $b_{n}={n+2-1\choose n}=n+1$? It's correct, but the way I always do it is by taking the derivative of $\frac{1}{1-x}=\sum x^k$, to get $\frac{1}{(1-x)^2}=\sum_{n=0}^\infty kx^{k-1}$ and from here we get $b_{n}=n+1$. $\endgroup$ – Julian Mejia May 16 at 16:01
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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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