Basically i'm having this recurrence relation which i don't know how to get the complexity of it by using the iterative method
$T(n) = \begin{cases} 0, & \text{if $n=0$} \\ 1, & \text{if $n=1$} \\ 3T(n-1)+ 4T(n-2), & \text{if $n >1$} \end{cases}$
Thanks in advance!
Solve: $x^2 - 3x - 4 = 0 \Rightarrow (x-4)(x+1) = 0 \Rightarrow x = 4, -1 \Rightarrow T_n = A\cdot 4^n + B\cdot (-1)^n$.
$T_0 = 0, T_1 = 1 \Rightarrow 0 = A+B, 1 = 4A-B \Rightarrow A = \dfrac{1}{5}, B = -\dfrac{1}{5} \Rightarrow T_n = \dfrac{4^n -(-1)^n}{5}$