I would like to solve following equation:
$$T(1) = 0$$
$$T(n) = 2T(n/2) + \Theta(n^2)$$
$$T(n)=\sum_{n=0}^{{\log_2n}}2^i \Theta\left(\frac{n^2}{2^{2i}}\right)$$
Now, I would like to compute it, but I am newbie at asymptotic. I only show you my approach (give good result), but I am not sure about corectness of computations:
$$\sum_{n=0}^{{\log_2n}}2^i \Theta\left(\frac{n^2}{2^{2i}}\right)=n^2\sum_{n=0}^{{\log_2n}} \Theta\left(\frac{2^i}{2^{2i}}\right)=n^2\sum_{n=0}^{{\log_2n}} \Theta\left(\frac{1}{2^{i}}\right)=n^2\cdot O(1) = \Theta(n^2)$$