<s>[INST] Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$ $a_0 = 3, a_1 = 7, a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$
I know I've got the wrong answer because my outputs don't match. So if someone could show me where I'm going wrong I would be super grateful.
Let $ y = \sum\limits_{n = 0}^{\infty} a_nx^n$
multiplying original formula through by $x^n$ I get
$a_nx^n = 3a_{n-1}x^n - 2a_{n-2}x^n$
sum over $n \geq 2$
$\sum\limits_{n \geq 2} a_nx^n = \sum\limits_{n \geq 2} 3a_{n-1}x^n - \sum\limits_{n \geq 2}2a_{n-2}x^n$
The left hand side is equal to $y - a_1 - a_0$
The first part of the right hand side
$\sum\limits_{n \geq 2} 3a_{n-1}x^n = 3x \sum\limits_{n \geq 2} a_{n-1}x^{n-1} = 3x(y-a_0)$
The second part of the right hand side
$\sum\limits_{n \geq 2}2a_{n-2}x^n = 2x^2\sum\limits_{n \geq 2}a_{n-2}x^{n-2} = 2x^2y$
So putting it all together and substituting for $a_0$ and $a_1$
$y - 10 = 3x(y-3) - 2x^2y$
all the $y$'s to one side
$y - 3xy + 2x^2y = -9x + 10 \Rightarrow y = \cfrac{-9x + 10}{(1-2x)(1-x)}$
after fraction decomposition I get
$y = \cfrac{11}{(1-2x)}-\cfrac{1}{(1-x)}$
Which then gives me
$a_n = 11 \cdot 2^n -1$
which is definitely wrong :(
 [/INST] Alternatively:
$$a_n = 3a_{n-1} - 2a_{n-2} \Rightarrow a_n-2a_{n-1}=a_{n-1}-2a_{n-2}.$$
Change: $b_n=a_n-2a_{n-1}, b_0=1$ to get:
$$b_n=b_{n-1} \Rightarrow b_n=1.$$
Hence:
$$a_n-2a_{n-1}=1, a_0=3 \Rightarrow a_n=2^{n+2}-1.$$
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