The function $f(z)$ can be expanded into two partial fractions $ f(z):=\frac{1}{\left( z-1\right) \left( z-2\right) }=\frac{1}{z-2}-\frac{1}{z-1}. $
We now expand each partial fraction into a geometric series. On $R_{2}$ these series are $ \begin{eqnarray*} \frac{1}{z-2} &=&\frac{1}{z\left( 1-2/z\right) }=\frac{1}{z} \sum_{n=0}^{\infty }\left( \frac{2}{z}\right) ^{n}\qquad \left\vert z\right\vert >2 \\ &=&\frac{1}{z}\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n}}=\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n+1}} \end{eqnarray*} $ and $ \begin{eqnarray*} \frac{1}{z-1} &=&\frac{1}{z\left( 1-1/z\right) } \\ &=&\frac{1}{z}\sum_{n=0}^{\infty }\left( \frac{1}{z}\right) ^{n}=\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}\qquad \left\vert z\right\vert >1. \end{eqnarray*} $
And so, the Laurent series is
$ \frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\frac{1}{ z^{n+1}}(2^{n}-1)\qquad \left\vert z\right\vert >2>1. $
On $R_{1}$, the two geometric series are $ \begin{eqnarray*} \frac{1}{z-2} &=&\frac{-1/2}{1-z/2}=\sum_{n=0}^{\infty }\left( -\frac{1}{2} \right) \left( \frac{z}{2}\right) ^{n}\qquad \left\vert z\right\vert <2 \\ &=&\sum_{n=0}^{\infty }-\frac{1}{2^{n+1}}z^{n} \end{eqnarray*} $
and
$ \begin{eqnarray*} \frac{1}{z-1} &=&\frac{1/z}{1-1/z}=\sum_{n=0}^{\infty }\frac{1}{z}\left( \frac{1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\ &=&\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}. \end{eqnarray*} $
We thus get the following Laurent series
$ \frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\left( - \frac{1}{2^{n+1}}z^{n}-\frac{1}{z^{n+1}}\right) \qquad 1<\left\vert z\right\vert <2. $