Finding a formula for a sequence 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 $1$?
 A: 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 $$a_n={3+(-1)^n\over 2}$$
A: 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
$$a_n=A\cdot (-1)^n+B\cdot 1^n=A\cdot (-1)^n+B \tag{1}$$
From
$$a_0=5=A+B$$
$$a_1=-2=-A+B$$
we have $B=\frac{3}{2}$ and $A=\frac{7}{2}$ or
$$a_n=\frac{7}{2}\cdot (-1)^n+\frac{3}{2} \tag{2}$$
A: 
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:
$$\sum_{n=0}^{\infty} a_nx^{n+1}+\sum_{n=0}^{\infty} a_{n+1}x^{n+1}=3\sum_{n=0}^{\infty} x^{n} \Rightarrow \\
xf(x)+f(x)-a_0=3\cdot \frac1{1-x} \Rightarrow \\
f(x)=\frac{5-2x}{(1-x)(1+x)}=\frac7{2(1+x)}+\frac3{2(1-x)}=\\
\frac72\sum_{n=0}^{\infty}(-x)^n+\frac32\sum_{n=0}^{\infty} x^n =\\
\sum_{n=0}^{\infty}\left[\frac72(-1)^n+\frac32\right]x^n \Rightarrow \\
a_n=\frac72(-1)^n+\frac32.$$
Can you solve the recurrence relation $a_n+a_{n+1}=3$, if $a_0=2, a_1=1$?
