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I have been trying to figure out this for quite some time but my basic mathematical understanding does not want to co-operate.

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Hint. One may write $$ X(z)=\frac53 \cdot \frac{1}{1-5\cdot z^{-1}} $$ then using the linearity of the $Z$-transform and using a table of common $Z$-transform pairs gives $$ \frac53 \cdot5^n \: u[n] $$ as being the signal, where $$ u : n \mapsto u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} $$ is the Heaviside step function.

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    Can you explain what happens to the 15? You first take the quotient ( 5/3) out and then eliminate z's to get ones.2017-01-22
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    I've written the denominator as $3z-15=3z \cdot \left(1-\frac{15}{3z}\right)=3z \cdot \left(1-5 z^{-1}\right)$ since $\frac{15}3=5$.2017-01-22