sequence with 4 multiple difference of previous two values

Question

tried the first few terms, they are 0, 4, 16, 48, 128, 320, 768, 1792, 4096, ... but can't figure out the general pattern behind it. Is there a general formula for $a_{n}$ in terms of n? Thanks for any help.

Answer 1

We need to solve the recurrence $$a_n - 4a_{n-1} + 4a_{n-2} = 0$$ with initial conditions $a_1 = 0,a_2=4$.

Let's proposed $a_n = r^n$ and substitute, we get $$ 0 = r^n - 4r^{n-1} + 4r^{n-2} = r^{n-2} \left[r^2 - 4r +4\right] = r^{n-2}(r-2)^2 $$ So $a_n = 2^n (A + Bn)$ for constants $A,B \in\mathbb{R}$. PLug in the initial conditions.