<s>[INST] Suddenly a wild Z-transform appears... I am attempting to complete the z transform of the following formula:
$$x(n) = \frac{1}{2}(n^2+n) u(n-1)$$
I got it into the summation form like so:
$$X(z) = \frac{3}{2} \sum_{n=1}^{\infty} n^2(\frac{1}{3} z^{-1})^2 + \frac{3}{2} \sum_{n=1}^{\infty} n(\frac{1}{3} z^{-1})^2$$
$$X(z) = ??? + 3\frac{\frac{1}{3}z^{-1}}{(1-\frac{1}{3}z^{-1})^2}$$
I'm not sure what to do about that first term there. It doesn't seem to fit any of the forms given in my Z-Transform table.
 [/INST] The crucial idea here is to use the fact that
$$
\sum_{n=1}^\infty na(n)z^{-n}=-z\frac{d}{dz}\sum_{n=1}^\infty a(n)z^{-n}.
$$
The Z-transform of your $x(n)$ will by definition be
$$
X(z)
=\sum_{n=-\infty}^\infty \frac{1}{2}(n^2+n)u(n-1)z^{-n}
=\frac{1}{2}\sum_{n=1}^\infty n^2z^{-n}+\frac{1}{2}\sum_{n=1}^\infty nz^{-n},
$$
where we can handle each of the two series separately.
We find that
$$
\sum_{n=1}^\infty nz^{-n}
=-z\frac{d}{dz}\sum_{n=1}^\infty z^{-n}
=-z\frac{d}{dz}\Big(\frac{z^{-1}}{1-z^{-1}}\Big)
=\frac{z}{(z-1)^2}
$$
and
$$
\sum_{n=1}^\infty n^2z^{-n}
=-z\frac{d}{dz}\sum_{n=1}^\infty nz^{-n}
=-z\frac{d}{dz}\bigg(\frac{z}{(z-1)^2}\bigg)
=\frac{z^2+z}{(z-1)^3}.
$$
Finally, this gives us
$$
X(z)
=\frac{1}{2}\bigg(\frac{z}{(z-1)^2}+\frac{z^2+z}{(z-1)^3}\bigg)
=\frac{z^2}{(z-1)^3}
=\frac{z^{-1}}{(1-z^{-1})^3}.
$$
</s>