<tex>\dfrac{ 1}{(1-z)^2}=(1-z)^{-2}=\sum\limits_{n=0}^{\infty} {-2\choose n}(-z)^n=\sum\limits_{n=0}^{\infty} (-1)^n{n+1\choose 1}(-z)^n=\sum\limits_{n=0}^{\infty}(n+1)z^n</tex>
<tex>=\sum\limits_{n=0}^{\infty} (-1)^n{n+1\choose 1}(-z)^n=\sum\limits_{n=0}^{\infty}(n+1)z^n</tex> <tex>G(z)=\dfrac{1/3}{(1-z)^2}+\dfrac{7/9}{1-z}-\dfrac{1/2}{1-2z}+\dfrac{7/18}{1-4z}=</tex> <tex>=\dfrac{1}{3}\sum\limits_{n=0}^{\infty} (n+1)z^n +\dfrac{7}{9}\sum\limits_{n=0}^{\infty} z^n - \dfrac{1}{2}\sum\limits_{n=0}^{\infty} 2^n z^n + \dfrac{7}{18}\sum\limits_{n=0}^{\infty} 4^n z^n</tex>
