Linear Algebra Recurrence equation problem

Given that x0=0 and x1=1 and for each number n we have the recurrence equation x(n+2)=3x(n+1)-2x(n) find a formula for x(n) in terms of n.

Everything in parentheses is subscript. How do I go about solving this question? I have been ripping out what is left of my hair for well over 30 mins and cant even get started ... A full solution with steps explained would be ideal.

Thanks!

If you know about the solution of linear recurrence equations, you assume that x(n) is of the
form x(n)=r^n so x(n+1)=r^(n+1) and x(n+2)=r^(n+2). Sub into the equation to get
r^(n+2)-3r^(n+1) +2r^n=0 and divide each term by r^n to give
r^2-3r+2=0 ans solving gives r=1 , r=2
The general solution is x(n)=A(1)^n+B(2^n)=A+B(2^n) for constants A and B
You know x(0)=0, so A+B=0 and x(1)=1 so A+2B=1 and you can find A and B

You need to study these carefully in your text book or on the Wikipedia
The solution is similar to solving a second order homogenius DE like
y" -3y'+2y=0

obviously this is 2^n -1 (n = 0,1,...)
0
1
3
7
15
31
63
127
255
511
1023

Just apply sigme from n = 0 to n = Say N on both sides of the equation and you get it.