Laurent Series Question: Summation Starts from $-\infty$?

Question

Consider $f(z)=\frac{2z-4}{z^2-4z+3}$. Find the Laurent series for $f(z)$ that converges when $|z-1|>2$.

My working:

\begin{align} f(z)&=\frac{1}{z-3}+\frac{1}{z-1} \\ &=\frac{1}{z-1}\left(\frac{1}{1-\frac{2}{z-1}}\right)+\frac{1}{z-1} \\ &=\sum_{n=0}^{\infty} \left(\frac{2^n}{(z-1)^{n+1}}\right)+\frac{1}{z-1} \end{align}

Now, the answer provided is, $$\sum_{n=-\infty}^{-1} c_n(z-1)^n, \ \ c_{-1}=2 \ \text{and} \ c_n=2^{-n-1} \ \ n\leq -2.$$ I do not understand in the slightest why this is. What am I missing?

Answer 1

They are same. But there seem to be typos. It is $c_{-1}$ and $n\leq -2$.

Answer 2

$(z-1)^{-k}=\frac{1}{(z-1)^k}$; your answers seem the same up to indexing