Give an asymptotic upper bound for $T(n) = 4T(\frac{n}{2}) + n^2log_2n$ Give an asymptotic upper bound for $$T(n) = 4T(\frac{n}{2}) + n^2log_2n$$ Using the recursion tree method, I obtain $$\sum_{i=0}^{log_2n - 1}  \frac{n}{2^i}log_2\frac{n}{2^i} + n^2\theta(1)$$
Not sure how to solve this summation.

From
$$
T\left(2^{\log_2 n}\right)=4T\left(2^{\log_2 \frac n2}\right)+n^2\log_2 n
$$
making $\mathbb{T}(\cdot) = T(2^{(\cdot)}),\ \ z = \log_2 n$ we have
$$
\mathbb{T}(z) = 4\mathbb{T}(z-1)+z 4^z
$$
This recurrence has the solution
$$
\mathbb{T}(z) = \frac 12 z(z+1)4^z+C_0 4^{z-1}
$$
and backtracking we have
$$
T(n) = C_1n^2+\frac 12(\log_2 n+1)n^2\log_2 n
$$
then 
$$
T(n) \approx \Theta\left((n\log_2 n)^2\right)
$$
