Ytukyg

Let $a_n$ be a sequence such that $a_0=5$ and $a_{n}+a_{n+1}=3$ for all $n$ greater than $0$.

I defined a function $f(x)$ whose coefficients are same of that mentioned sequence.
I was able to get $$f(x)=displaystyle sum_{n=0}^infty [ 5 (-1)^n x^n + 3 (-1)^n x^{n+1} + 3 x^{2n+1} ] $$
But couldn't make it far.
How do you find a formula for $a_n$ given that $a_1$ is just $1$?

Define $b_n=a_n-1.5$. Therefore$$b_n+b_{n+1}=0$$which means that$$b_{n+1}=b_1(-1)^n$$or $$a_n=1.5+k(-1)^n$$with $a_1=1$ we have $$a_1=1=1.5+k(-1)to k= 0.5$$therefore $$a_n={3+(-1)^nover 2}$$

Another way is to use characteristic polynomial method (here is another example). Given
$$a_n+a_{n+1}=3$$
$$a_{n-1}+a_n=3$$
we have $a_{n+1}-a_{n-1}=0$, with characteristic polynomial
$$x^2-1=0$$
with solutions $-1,1$ and general term
$$a_n=Acdot (-1)^n+Bcdot 1^n=Acdot (-1)^n+B tag{1}$$
From
$$a_0=5=A+B$$
$$a_1=-2=-A+B$$
we have $B=frac{3}{2}$ and $A=frac{7}{2}$ or
$$a_n=frac{7}{2}cdot (-1)^n+frac{3}{2} tag{2}$$

Let an be a sequence such that $a_0=5$ and $a_n+a_{n+1}=3$ for all $n$ greater than $0$.

First note: $a_0+a_1=3 Rightarrow 5+a_1=3 Rightarrow a_1=-2$.

It looks you are trying to use the generating function $f(x)=sum_{n=0}^{infty} a_nx^n$. Here are the steps:
$$sum_{n=0}^{infty} a_nx^{n+1}+sum_{n=0}^{infty} a_{n+1}x^{n+1}=3sum_{n=0}^{infty} x^{n} Rightarrow \
xf(x)+f(x)-a_0=3cdot frac1{1-x} Rightarrow \
f(x)=frac{5-2x}{(1-x)(1+x)}=frac7{2(1+x)}+frac3{2(1-x)}=\
frac72sum_{n=0}^{infty}(-x)^n+frac32sum_{n=0}^{infty} x^n =\
sum_{n=0}^{infty}left[frac72(-1)^n+frac32right]x^n Rightarrow \
a_n=frac72(-1)^n+frac32.$$

Can you solve the recurrence relation $a_n+a_{n+1}=3$, if $a_0=2, a_1=1$?