Partial fraction of $A(x)=\frac{x^2+x+1}{(1-x)^3}$

Question

I was trying to get a formula for some sequence but I'm stuck at last part and I really want to know if there's a general way of solving it: What is the numerator of partial fractions of …> $$A(x)=\frac{x^2+x+1}{(1-x)^3}$$ Appreciate any help!

Answer 1

$$-\frac{x^2+x+1}{(x-1)^3}=-\frac{x^2-2x+1+3x-3+3}{(x-1)^3}=-\frac{1}{x-1}-\frac{3}{(x-1)^2}-\frac{3}{(x-1)^3}$$

Answer 2

HINT: Write $A(x)$ as $$\frac{x^2+x+1}{(1-x)^3} = \frac{B}{1-x}+\frac{C}{(1-x)^2}+\frac{D}{(1-x)^3}$$ and find $B$, $C$ and $D$.

Answer 3

$$\frac{x^2+x+1}{(1-x)^3}=\frac{x^2-2x+1+3x}{(1-x)^3}=\frac{(x-1)^2+3(x-1)+3}{(1-x)^3}=\frac{(x-1)^2}{(1-x)^3}+\frac{3(x-1)}{(1-x)^3}+\frac{3}{(1-x)^3}$$ $$=-\frac{(x-1)^2}{(x-1)^3}-\frac{3(x-1)}{(x-1)^3}-\frac{3}{(x-1)^3}$$

and then complete it

Answer 4

Actually, you do not need to perform a partial fraction decomposition.

By stars and bars $$ \frac{1}{(1-x)^4} = \sum_{n\geq 0}\binom{n+3}{3}x^n \tag{1}$$ and since $\frac{1+x+x^2}{(1-x)^3}=\frac{1-x^3}{(1-x)^4}$ it is enough to multiply the RHS of $(1)$ by $1-x^3$ to get $$ \frac{1+x+x^2}{(1-x)^3}=\sum_{n\geq 0}\left[\binom{n+3}{3}-\binom{n}{3}\right]x^n=\sum_{n\geq 0}\frac{3n^2+3n+2}{2}\,x^n.\tag{2} $$