solving recurrence relation with Z-transform (LTI and causal)
With such simple recursions it's often easiest (and instructive) to actually go through the recursion and find a pattern. With $y[0]=20$ and using $c=35$ we get

$$\begin{align}y[1]&amp;=c+\frac{y[0]}{2}\\y[2]&amp;=c+\frac{c}{2}+\frac{y[0]}{4}\\y[3]&amp;=c+\frac{c}{2}+\frac{c}{4}+\frac{y[0]}{8}\\\vdots\end{align}$$

So we can conclude

$$\begin{align}y[n]&amp;=c\left[1+\frac12+\frac14+\ldots+\left(\frac12\right)^{n-1}\right]+y[0]\left(\frac12\right)^n\\&amp;=c\sum_{k=0}^{n-1}\left(\frac12\right)^k+y[0]\left(\frac12\right)^n\\&amp;=c\cdot\frac{1-\left(\frac12\right)^{n}}{1-\frac12}+y[0]\left(\frac12\right)^n\\&amp;=2c+\big(y[0]-2c\big)\left(\frac12\right)^n,\qquad n&gt;0\end{align}$$

which is the same expression you obtained.