Iteration

$$T(n) \;=\; n^3 \;+\; 2T(n/2)$$

$$=\; n^3 \;+\; 2[(n/2)^3 \;+\; 2T(n/4)]$$

$$=\; n^3 \;+\; \frac{n^3}{2^2} \;+\; 4T(n/4)$$

$$=\; n^3 \;+\; \frac{n^3}{2^2} \;+\; 4[(n/4)^3 \;+\; 2T(n/8)]$$

$$=\; n^3 \;+\; \frac{n^3}{2^2} \;+\; \frac{n^3}{4^2} \;+\; 8T(n/8)$$

These are terms i=0, i=1, i=2, i=3

$$=\; n^3 \;+\; \frac{n^3}{2^2} \;+\; \frac{n^3}{4^2} \;+\; \frac{n^3}{8^2} \;+\; ... \;+\; 2^i T(\frac{n}{2^i})$$

Terminates when n/2ⁱ = 2, T(2) = $\Theta(1)$.

n = 2ⁱ⁺¹, lgn = i+1, i = lgn - 1

$$=\; \sum_{i=0}^{lgn - 2}\frac{n^3}{(2^i)^2} \;+\; 2^{lgn - 1} T(2)$$

$$=\; n^3 \sum_{i=0}^{lgn - 2} \; (1/4)^i \;+\; 1/2 n \Theta(1)$$

$$=\; n^3 (\frac{(1/4)^{lgn - 1} - 1}{1/4 - 1}) \;+\; 1/2 \Theta(n)$$

$$=\; n^3 (\frac{1 - (2^{-2})^{lgn - 1}}{3/4}) \;+\; \Theta(n)$$

Note that (1/4)^{lgn - 1} = 2^{-2 lg n + 2}

= = 4 n^-2

$$=\; 4/3n^3 (1 - 4n^{-2}) \;+\; \Theta(n)$$

$$=\; 4/3n^3 - 16/3n + \Theta(n)$$

$$=\; \Theta(n^3)$$