Solution Techniques

Guess: T(n) = O(2ⁿ), which means T(n) $\leq$ c2ⁿ

$$T(n) \;\;=\;\; T(n-1) \;+\; T(n-2) \;+\; \Theta(1)$$

$$<= \;\; c 2^{n-1} \;+\; c2^{n-2} \;+\; \Theta(1)$$

$$=\;\; 1/2c2^n + 1/4c2^n + \Theta(1)$$

$$=\;\; 3/4c2^n + \Theta(1)$$

$$<=\;\; c2^n$$

If $1/4c2^n \geq \Theta(1)$ or $c \geq 4\Theta(1)/2^n$.

True for sufficiently large n.

Boundary: T(0) = 1 $\leq$ c2⁰ = c, c$\geq$1

Actually, T(n) = .