# Origin under $z$ transformation of $\frac{6}{z-2}$

I am trying to find the origin of $\frac{6}{z-2}$ under the $z$ transformation, i.e. find some $f(k)$ s.t. $$f(k)\overset{z}{\rightarrow}F(z)=\frac{6}{z-2}$$

According to a formula I have the origin for $\frac{A}{(z-a)^{m+1}}$ is $$A\frac{k-m}{k}\binom{k}{m}a^{k-m-1}$$ when $k\geq m+1$ and $0$ otherwise.

This gave me the solution $6\cdot2^{k-1}$when $k\geq1$ and $0$ otherwise, but the solution say the answer is $6\cdot2^{k}$ (when $k\geq1$ and $0$ otherwise).

Do I have a mistake ? if so, what is the source of the mistake (e.g. wrong formula, correct formula so I must have a calculation error of some sort etc')

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Here the solution you say you got seems exactly to be the solution the answer says. Did you mean to give two different answers and ask which is right? – coffeemath Oct 20 '12 at 19:21
@coffeemath - thanks for pointing that out, it was a typo. the solution say it is $6\cdot 2^k$ but I have it as $6\cdot 2^{k-1}$ – Belgi Oct 20 '12 at 19:26
If you just match things, A=6,a=2,m=0 and plug in you get precisely what you wrote as your answer. So if the "answer" in the book is not that, it means the formula must be wrong. – coffeemath Oct 20 '12 at 19:27
@coffeemath - either it was a typo in the book... – Belgi Oct 20 '12 at 19:31
Your answer must be right, since if $m>0$ there would be a factor of $k$ from the "k choose m". In my experience the backs of books often have minor errors. If we put m=0 in the exponent k-m-1 it ios certainly k-1, and if someone thought it was k they would have to think that m=1, which would give the extra factor of k. But the answer quoted has no factor of k in the pretransform. So I think you're right. – coffeemath Oct 20 '12 at 19:32