$$f(x)=\frac{2x+3}{x^2 -4x+5}$$ on $x=2$.
My solution:
$t=x-2 $ => $x=t+2$ , we get: $f(t)=\frac{2t+7}{t^2+1}$ on $t=0$.
then: $(2t+7)\sum_{n=0}^{\infty } {(-t^2)^n} = (2t+7)\sum_{n=0}^{\infty }{(-1)^nt^{2n}}$.
The solution MUST BE of the form: $\sum_{n=0}^{\infty} g(n)(x-2)^{h(n)}$.
Any help on how to continue?
Thanks.
Your approach looks good. In what form do you want/need it?
You could rewrite: $$(2t+7) \sum_{n=0}^{\infty} (-1)^nt^{2n} = \sum_{n=0}^{\infty} (-1)^n(2t+7)t^{2n} = \sum_{n=0}^{\infty} (-1)^n\left( 2t^{2n+1}+7t^{2n} \right)$$ And substitute back to $x$: $$= \sum_{n=0}^{\infty} (-1)^n\left( 2(x-2)^{2n+1}+7(x-2)^{2n} \right)$$
If you really want it in the standard form $$\sum_{n=0}^{\infty} a_n (x-2)^n$$ then you can group the powers of $(x-2)$ but you'll need some adjustments to get the coefficients of the even and odd powers right. The odd powers have coefficient $2$ and $(1-(-1)^n)/2$ is $1$ for $n$ odd and $0$ for $n$ even; so the contribution of the odd terms can be put into the factor: $$\frac{1-(-1)^n}{2}2$$ In a similar way, the contribution of the even terms can be put into the factor: $$\frac{1-(-1)^{n+1}}{2}7$$ That would give the following factor which can be simplified: $$\frac{1-(-1)^n}{2}2+\frac{1-(-1)^{n+1}}{2}7 = \tfrac{1}{2} \left( 9+5 (-1)^n \right) $$ Note however that we still didn't take the factor $(-1)^n$ into account; this now has to alternate every two terms (1, 1, -1, -1, ...). You can do this with sines and cosines or e.g. with a factor: $$(-1)^{\tfrac{1}{4}\left( 2n-1+(-1)^n \right)}$$ to give: $$\sum_{n=0}^{\infty} (-1)^{\tfrac{1}{4}\left( 2n-1+(-1)^n \right)}\tfrac{1}{2} \left( 9+5 (-1)^n \right) (x-2)^n$$