Closed form of $\mathcal Z$-transform : decomposition signal $x(n)$
Or you can proceed this way: \begin{align}\mathcal Z\left\{x(n)\right\}= X(z) &amp;= \sum_{n=-\infty}^\infty x(n)z^{-n}\\
&amp; = 0 + \sum_{n=0}^{2N} \left|n - N\right|z^{-n} + 0\\
&amp; = \sum_{n=0}^{N} \left(-n + N\right)z^{-n} + \sum_{n=N}^{2N} \left(n - N\right)z^{-n}\\
&amp; = N\sum_{n=0}^{N}z^{-n} - N\sum_{n=N}^{2N}z^{-n} - \sum_{n=0}^{N} nz^{-n} + \sum_{n=N}^{2N} nz^{-n} 
\end{align}

And changing the boundaries for the second and fourth term, putting $k = n - N$, you get :

$$X(z) = N\sum_{n=0}^{N}z^{-n} - Nz^{-N}\sum_{k=0}^{N}z^{-k} - \sum_{n=0}^{N} nz^{-n} + z^{-N}\sum_{k=0}^{N} kz^{-k} $$

You can continue from there.