Solving a homogeneous recurrence relation

Question

How to solve $$a_n=4a_{n-1}-4a_{n-2}$$ $a_0=6$ and $a_1=8$

I found first few terms as $$6,8,8,0,... $$ But I don't know how to proceed.

Answer 1

Make its characteristic equation: $$x^2-4x+4=0$$ Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$ Just substitute for initial conditions and you get $A=6$ and $B=-2$

So, $$a_n=(6-2n)\cdot2^n$$