Hi. I have this problem.

Let $\displaystyle H(x)=\sum_{n\ge 0}{h_nx^n}$ and $\displaystyle H(x)=\frac{1-2x+2x^2}{(1-x)^2(1-2x)}$

a)Find the recurrence relation the coefficients $\displaystyle h_n$ satisfy. Be sure to state the initial values.

Okay, this is what I did so far. I went about finding the partial fractions for this $\displaystyle H(x)=\frac{1-2x+2x^2}{(1-x)^2(1-2x)}$. And assuming I did it correctly (I checked it on a calculator so it should be correct)... I ended up with this

$\displaystyle H(x)=\frac{2}{1-2x}-\frac{1}{(1-x)^2}$

Then my nxt step was to place it into $\displaystyle H(x)=\sum_{n\ge 0}{h_nx^n}$ form. So I get,

$\displaystyle H(x)=2\sum_{n\ge 0}{(2x)^n-\sum_{n\ge 0}{(n+1)x^n}$ by geometric series.

$\displaystyle =2{\sum_{n\ge 0}{(2^n-\frac{1}{2}(n+1)) x^n$

so $\displaystyle h_n=2(2^n-\frac{1}{2}(n+1))$

It doesn't look right to me. Because I don't know what the initial values are. $\displaystyle h_0=\frac{1}{2}$?

b) Find an explicit expression for $\displaystyle h_n$, $\displaystyle n\ge0$

How do I do this one? What would be the difference between finding the recurrence relation??

Thanks for your help.