Z transform of $\sum_{k=0}^{n}3^{k}$

Question

My task is to calculate z transform of signal $x[n]=\sum\limits_{k=0}^{n}3^{k}$ ?

By definition, $$ \begin{align} X(z) &= \sum\limits_{n=-\infty}^{n=\infty}x[n]z^{-n} \\ &= \sum\limits_{n=-\infty}^{n=\infty}\left(\sum\limits_{k=0}^{n}3^{k}\right)z^{-n} \\ &= \sum\limits_{n=-\infty}^{n=\infty}(3^{0}+3^{1}+3^{2}+...+3^{n})z^{-n} \\ \end{align}$$

but I don't know how to deal with this sum. Any suggestion?

Answer 1

Hint. Observe that, using $$ \sum_{k=0}^{n}a^{k}=\frac{a^{n+1}-1}{a-1}, \quad a \neq 1, $$ you get $$ \sum_{k=0}^{n}3^{k}=\frac{3^{n+1}-1}2 $$ then $$ X(z)=\sum_{n=-\infty}^{\infty}(3^{0}+3^{1}+3^{2}+...+3^{n})z^{-n}=\sum_{n=-\infty}^{\infty}\left(\frac{3^{n+1}-1}2\right)z^{-n} $$ which you may simplify.