Modified Z-transform

Question

Can somebody tell me where I do a mistake in derivation of modified (or advanced) Z-transform of digital parabolic sequence i. e $$f(k) = (k\cdot T)^2,$$ where $T$ is the sampling period and $k$ is number of a sample. I started from definition of modified Z-transform $$\mathcal{Z}\{f(k),m\} = z^{-1}\cdot\sum_{k=0}^{\infty}(k\cdot T + m\cdot T)^2\cdot z^{-k},$$ where $m$ is a fraction of sampling period. Then I evaluated the power and used linearity of Z-transform. Thus I got $$ z^{-1}\cdot\left(\sum_{k=0}^{\infty}(k\cdot T)^2\cdot z^{-k} + \sum_{k=0}^{\infty}(2\cdot k\cdot T\cdot m \cdot T) \cdot z^{-k} + \sum_{k=0}^{\infty}(m\cdot T)^2\cdot z^{-k}\right).$$ Then I used Z-transforms of elementary sequences and I got $$z^{-1}\cdot\left(\frac{T^2\cdot z \cdot (z+1)}{(z-1)^3}+2\cdot m\cdot T\frac{T\cdot z}{(z-1)^2}+(m\cdot T)^2\cdot \frac{z}{z-1}\right).$$ This result is unfortunately slightly different from the result in modified Z-transform table: $$T^2\cdot\left(\frac{m}{z-1}-\frac{2\cdot m+1}{(z-1)^2}+\frac{2}{(z-1)^3}\right).$$ But when I used MATLAB command ztrans, I got the same result as me. I am confused. Thanks for any suggestions.