Easy Z-Transform doesn't match in Wolfram/Matlab I have this function:
$$g(k)=2*(\frac14)^{k-1}+(\frac18)^k$$
I calculate the Z-transform in this way, applying the delay property: $Z[x(k-1)]=z^{-1}X(z)$
$$G(z)=\frac{8}{4z-1}+\frac{8z}{8z-1}$$
The problem is that both Wolfram and Matlab solved it in this way:
$$G(z)=\frac{32z}{4z-1}+\frac{8z}{8z-1}$$
They are considering:
$$(\frac14)^{k-1}=(\frac14)^{k}*(\frac14)^{-1}$$
How can it be possible? The delay property is not applicable here, or they are wrong?
 A: Z-transform can be two-sided (bilateral), namely
$$\sum_{k=-\infty}^\infty  g(k) z^{-k} \tag1$$
In this case, delay property holds as you stated:
$$\sum_{k=-\infty}^\infty  g(k-1) z^{-k} =\sum_{k=-\infty}^\infty  g(k) z^{-k-1}
= z^{-1}\sum_{k=-\infty}^\infty  g(k) z^{-k} $$
However, Maple et al calculate one-sided Z-transform: 
$$\sum_{k=0}^\infty  g(k) z^{-k} \tag2$$
For this transform the delay property is different,  since there is no $g(-1)$  in (2). We have the following: 
$$ z^{-1}\sum_{k=0}^\infty  g(k) z^{-k}  = \sum_{k=0}^\infty  g(k) z^{-k-1} 
= \sum_{k=1}^\infty  g(k-1) z^{-k}  $$
Now, if we are dealing with $g(k) = 2/4^{k-1}$, then $g(k-1)=4g(k)$, hence
$$
z^{-1}\sum_{k=0}^\infty  g(k) z^{-k} = 4 \sum_{k=1}^\infty  g(k) z^{-k} 
= 4 \sum_{k=0}^\infty  g(k) z^{-k}  - 4g(0)
$$
Solve:
$$\sum_{k=0}^\infty  g(k) z^{-k} = \frac{4g(0)}{4-z^{-1}} = \frac{32z}{4z-1}$$
in agreement with Maple et al. 

By the way, it is easier not to use any kind of delay property and calculate directly using the sum of geometric series: 
$$\sum_{k=0}^\infty ar^k = \frac{a}{1-r}$$
