I have the equation $T(n) = 4T(n/2) + n + log(n)$ for $n\ge2$.
I am considering the case where $n=2^k$
I have come to the conclusion that $T(n)$ follows the following formula:
$\begin{align*}T(n) = [n+\log(n)] & + 4[(n/2)+\log(n/2)] + 4^2[(n/2^2) + \log(n/2^2)] + \ldots \\ &+ 4^{k-1}[(n/2^{k-1}) + \log(n/2^{k-1}] + 4^k[T(0)]\end{align*}$
Based on the formula that for $n\ge b$: $\begin{align*}T(n) = aT(n/b) + f(n)\end{align*}$
$T(n)$ goes to
$\begin{align*}T(n) = f(n) + a\cdot f(n/b) + \ldots + a^i\cdot f(n/b^i) + \ldots + a^{k-1}\cdot f(n/b^{k-1}) + a^k\cdot T(0)\end{align*}$
However, I don't know how to reduce this function. I need determine what the rate of growth, i.e. the order of the function is, in terms of $n$. This is problem 13 in the pdf here.
That pdf also contains the formulas I've used to get thus far.