Need more help on z transform(pls check if this is correct)

I. Find the z transform and ROC of each of the ff sequence

1.$\displaystyle x(n) = 2\delta{n} + 3(\frac{1}{2})^{n}u[n] - (1/4)^{n}u(n)$

$\displaystyle X(z) = 2(1) + 3\frac{1}{ 1-\frac{1}{2}z^{-1} } - \frac{1}{1-\frac{1}{4}z^{-1} }$

ROC: All z, 1/2 <z and 1/4 < z

the total ROC is z > 1/2

II. Use the Z transform to perform the convolution of the following sequence

$\displaystyle x[n] = 3^{n}u(-n)$

since $\displaystyle x(-n)$'s z transform is $\displaystyle X(z^{-1})$

$\displaystyle X_1(Z) = \frac{1}{1-3z}$

$\displaystyle h[n] = (0.5)^{n}u(n)$

$\displaystyle X_2(Z) = \frac{1}{1-\frac{1}{2}z^-1}$

$\displaystyle x[n]*h[n]$ is equivalent to $\displaystyle X_1(Z)X_2(Z)$

$\displaystyle X_1(Z)X_2(Z) =\frac{1}{(1-3z)(1-\frac{1}{2}z^-1)}$

using wolfram alpha to solve partial fractions

$\displaystyle \frac{2z}{(1-3z)(2z-1)}$

$\displaystyle \frac{2}{3z-1} - \frac{2}{2z-1}$

then simplify

$\displaystyle \frac{ 2z^{-1}\frac{1}{3} }{ 1-\frac{1}{3}z^{-1} } - \frac{ 2z^{-1}\frac{1}{2} }{1-\frac{1}{2}z^{-1} }$

the region of convergence is 1/3 < z and 1/2 < z

then the total ROC is 1/2

the inverse z transform is : using the time shifting property z^-1X(z) = u(n-1)

$\displaystyle \frac{2}{3}(\frac{1}{3})^{n}u(n-1) -(\frac{1}{2})^{n}u(n-1)$

part III. Find the causal signal x(n) fo the following z transforms

$\displaystyle X(Z) = \frac{ 1-z^{-1} }{(1+z^{-1})(1+\frac{1}{2}z^{-1})}$

using wolfram alpha

$\displaystyle \frac{2(z-1)z}{z+1}{2z+1}$

$\displaystyle \frac{3}{2z+1}-\frac{4}{z+1}+1$

Making it into a z transform

$\displaystyle \frac{\frac{1}{2}3z^{-1}}{1-(-\frac{1}{2})z^{-1}} -\frac{4z^{-1}}{(1-(-z^{-1})} +1$

the region of convergence is

-1/2 < z, -1 < z, and the entire plane of z

the total region of convergence is the entire plane of z

the inverse z transform is

$\displaystyle x(n) = \frac{3}{2}(-\frac{1}{2})^{n}u(n-1) - 4(-1)^{n}u(n-1)$