Find coefficient of $x^{n}$ in $\frac{1}{(1-x)^3(1+x)^2}$

Question

I want to find a general expression for $x^{n}$ in $\frac{1}{(1-x)^3(1+x)^2}$.

I have tried to use partial fraction but that got complicated and I got stuck.

Any help would be appreciated.

Note: I would need a closed form expression

Answer 1

$\mathbf{Hint:} $

$$\frac{1}{(1-x)^3(1+x)^2}$$ $$=\frac{1}{(1-x^2)^2(1-x)}$$ $$= \left(\sum_{r=0} \binom{r+1}{r}.x^{2r}\right).\left(\sum_{r=0} x^{r}\right)$$ Can you take it from here

Answer 2

More generally, consider $\dfrac{1}{(1-x)^a(1+x)^b}$.

From the generalized binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem#Generalizations),

$(1-x)^{-s} =\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^k =\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k $.

Therefore $(1-x)^{-a} =\sum_{k=0}^{\infty} \binom{a+k-1}{a-1}x^k $ and $(1+x)^{-b} =\sum_{j=0}^{\infty} \binom{b+j-1}{b-1}(-1)^jx^k $.

Using the standard Cauchy product,

$(1-x)^{-a}(1+x)^{-b} =\sum_{n=0}^{\infty}x^n\sum_{m=0}^n \binom{a+m-1}{a-1}\binom{b+n-m-1}{b-1}(-1)^{n-m} $, so the coefficient of $x^n$ is $\sum_{m=0}^n \binom{a+m-1}{a-1}\binom{b+n-m-1}{b-1}(-1)^{n-m} $.

As to whether this can be simplified, I don't know.

Maybe something Vandermondish.

Answer 3

Lord Shark is right, do the partial fractions \begin{eqnarray*} \frac{1}{(1+x)^2 (1-x)^3} = \frac{3/16}{1+x}+\frac{1/8}{(1+x)^2}+\frac{3/16}{1-x}+\frac{1/4}{(1-x)^2}+\frac{1/4}{(1-x)^3}. \\ \end{eqnarray*} Now each of these terms can be expanded geometrically \begin{eqnarray*}\frac{1}{(1+x)^2 (1-x)^3} = \sum_{n=0}^{\infty} \left( \frac{3}{16}(-1)^n+\frac{1}{8}(-1)^n (n+1) +\frac{3}{16}+\frac{1}{4}(n+1)+\frac{1}{8}(n+1)(n+2) \right)x^n. \\ \end{eqnarray*}

Answer 4

By stars and bars $$ \frac{1}{(1-x)^2}=\sum_{n\geq 0}(n+1)x^n \tag{1}$$ and by replacing $x$ with $x^2$ $$ \frac{1}{(1-x^2)^2}=\sum_{n\geq 0}(n+1)x^{2n} \tag{2}$$ then by multiplying both sides by $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$ we have $$ \frac{1}{(1-x)(1-x^2)^2} = \sum_{m\geq 0}x^m \sum_{\substack{0\leq n\leq m\\n\text{ even}}}\left(\frac{n}{2}+1\right)\tag{3} $$ so the coefficient of $x^m$ in the LHS of $(3)$ is $\frac{(m+2)(m+4)}{8}$ is $m$ is even and $\frac{(m+1)(m+3)}{8}$ if $m$ is odd.