# 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:

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

which gives me:

$$20+2-1 \choose 20$$ $$-$$ $$7 \choose 1$$ $$15+2-1 \choose 15$$+$$7 \choose 210+2-1 \choose 10$$-$$7 \choose 345+2-1 \choose 5$$+$$7 \choose 40+2-1 \choose 0$$

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?

• 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 $b_{n}=n+1$. – Julian Mejia May 16 at 16:01

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 $$5\cdot (-35)=-175$$.