# Solve $a_n = 4a_{n-1}-4a_{n-2}$ with $a_0= 0$ and $a_1= 2$

Consider the linear recurrence relation $a_n = 4a_{n-1}-4a_{n-2}$. Find the solution of this recurrence relation for the initial conditions $a_0= 0$ and $a_1= 2$.

So I went through the motions and triangularized it got the answer $$a_n=1/2(2^n-2(2^n)a_1 - 4(2^n)a_o$$ But I know that that is wrong and the correct answer is $n*2^n$ But it just doesn't work for me. 4 -4 1 0 is the matrix I start with and I find it's eigenvalue to be only 2 and null space{(2 1)^t} so I extend it and to get the triangular matrix and come to the same answer you see above which is wrong.

You are solving $$a_n - 4a_{n-1} + 4a_{n-2}$$ and assuming $a_n = c^n$ you end up with $c^2 - 4c + 4 = 0$, so $c_\pm = 2$. Hence, indeed, the general solution would be $a_n = A2^n + Bn2^n$ and substitution of the initial conditions will determine the corresponding coefficients $A,B \in \mathbb{R}$.
Alternatively, express as: $$a_n-2a_{n-1}=2(a_{n-1}-2a_{n-2}).$$ Let $b_n=a_n-2a_{n-1}, b_1=2$. Then: $$b_n=2b_{n-1} \Rightarrow b_n=2^n.$$ So: $$a_n-2a_{n-1}=2^n.$$ Can you finish?
Hint: divide by $2^n$ and denote $c_n=\frac{a_n}{2^n}$.