Draw the recursion tree for $T(n) = 4T(\lfloor n/2 \rfloor) + cn$, where $c$ is a constant, and provide a tight asymptotic bound on its solution. Verify your answer with the substitution method.

Let's ignore the floor. Previous exercises illustrate how this can be handled.

The depth of the tree is $\lg{n}$, each level is $2^icn$ and there are $4^{\lg{n}} = n^2$ leaves.

Thus:

$$ T(n) = \sum_{i=0}^{\lg{n}- 1}2^icn + \Theta(n^2) = cn \sum_{i=0}^{\lg{n}-1}2^i + \Theta(n^2) = cn \frac{2^{\lg{n}} - 1}{2 - 1} + \Theta(n^2) = \Theta(n^2) $$

Let's guess $T(n) \le cn^2 + 2cn$:

$$ T(n) \le 4c(n/2)^2 + 2cn/2 + cn \le cn^2 + 2cn $$

Let's guess $T(n) \ge cn^2 + 2cn$:

$$ T(n) \ge 4c(n/2)^2 + 2cn/2 + cn \ge cn^2 + 2cn $$