Inverse Z-transform of $\frac{10}{z-5}-\frac{2}{z-1}$

Question

I must calculate the inverse z-transform of $\frac{10}{z-5}-\frac{2}{z-1}$. I decided to use the known formula $H(n-1)a^n\rightarrow \frac{a}{z-a}$, where $H(n)$ is the heaviside signal. I finally get this result: $2H(n-1)5^n-2H(n-1)$, but my professor's result would be $2H(n)(5^{n}-1)$. Can you explain me where's the matter please? Thankyou!

Answer 1

HINT: For $n>0$ both expressions are obviously identical. You seem to worry about $n=0$. Sit down and determine the value of both expressions at $n=0$ and you'll see that everything is fine.