solving recurrence relation with Z-transform (LTI and causal) I just wanted to doublecheck answers for my sanity's sake (exam next week)

problem statement

recurrence relation, solve it

$y[n+1]= 35 + y[n]*0.5$

according to my teacher it will be such that the input sequence must be causal, so that we have re-arranged form such as 

$y[n+1]-  y[n]*0.5= 35u[n]$
,where $y[0]=20$, and  causal input

My understanding is that we should go as follows


get $X(z)$
find what is $Y(z)$
then we can maybe do inverse Z-transform to get the y[k] output sequence in explicit form sequence
typically the most difficult phase seems to be the inverse Z transform so it can be tricky, but we have a table, and our teacher taught us to either use 


z-table
long division
partial fraction expansion and z-table





Firstly, Z-transform both sides

$Z(y[n+1])-0.5Z(y[n])=35*Z(u[n])$

according to the z-table we receive

$z*Y(z)-z*y[0]-0.5*Y(z)=35*\frac{z}{z-1}$

combine  terms and re-factor and emplace y[0]=20

$Y(z)(z-0.5)=35*\frac{z}{z-1}+z*20$

$Y(z)=35*\frac{z}{(z-1)(z-0.5)}+20*\frac{z}{z-0.5}$

We designate R(z) for the purposes of partial fraction expansion, because it is not found in the z-table

$R(z)=\frac{z}{(z-1)(z-0.5)}$

however the rightsided term is found directly in the z-table
$Z^{-1}(20*\frac{z}{z-0.5}) = 20* (0.5^k)$

start partial fraction expansions
$R(z)/z = \frac{1}{(z-1)(z-0.5)}=\frac{a}{(z-1)}+\frac{b}{(z-0.5)}$, then we find that, $a=2,b=-2$

$R(z)=2*\frac{z}{(z-1)}-2*\frac{z}{(z-0.5)}$

$Y(z)=35*(2*\frac{z}{z-1}-2*\frac{z}{z-0.5})+20*(\frac{z}{z-0.5})$

apply inverse Z transform

$y[k]=20*(0.5)^k+70*u[k]-70*0.5^k$

$y[k]=u[k]*(70-50*0.5^k)$

tabulating some results


k=-1; y[-1]=0
k=0; y[0]= 20
k=1; y1= 45
k=2; y[2]= 115/2
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.