# Solve a recurrence relation with two recursion calls using the iteration method [duplicate]

$$T(n) = \begin{cases} 0, & \text{if n=0} \\ 1, & \text{if n=1} \\ 3T(n-1)+ 4T(n-2), & \text{if n >1} \end{cases}$$
Hint: Let $S(n) = T(n+1) + T(n)$. We have $S(0) = 1$, and for $n > 0$, $$S(n) = T(n+1) + T(n) = [3T(n) + 4T(n-1)] + T(n) = 4(T(n) + T(n-1)) = 4S(n-1).$$ Can you take it from here?