How to tackle a recurrence that contains the sum of all previous elements? Say I have the following recurrence:
$$T(n) = n + T\left(\frac{n}{2}\right) + n + T\left(\frac{n}{4}\right) + n + T\left(\frac{n}{8}\right) + \cdots +n + T\left(\frac{n}{n}\right) $$
where $n = 2^k$, $k \in \mathbb{N} $ and $T(1) = 1$.
simplified to:
$$T(n) = n \log_2n + \sum_{i=1}^{\log_2n}T\left(\frac{n}{2^i}\right)  $$
The Master's theorem is not applicable; neither is the Akra-Bazzi method since $k = \log_2$ is not a constant. 
What strategy can I use to find a closed form solution? I have a feeling that the closed form is $T(n) = \sum_{i=0}^{\log_2n}\left[j\frac{n}{2^i}  \log_2 \left(\frac{n}{2^i} \right)\right] + 1 $ where $j = \max\left(1, 2^{i-1}\right)$ but would like a proof.

I'd just start with $T(1)$ and look for a pattern:
$$T(2^1) = 1 \cdot 2^1 + 2^{1-1}T(1)$$
$$T(2^2) = 2\cdot 2^2 + 1\cdot 2^1 + (2^{2-1}) T(1)$$
$$T(2^3) = 3\cdot 2^3 + 2\cdot 2^2 + 2 \cdot 1 \cdot 2^1 + 2^{3-1} T(1)$$
$$T(2^4) = 4\cdot 2^4 + 3\cdot 2^3 + 2 \cdot 2 \cdot 2^2 + 4 \cdot 1 \cdot 2^1 + 2^{4-1} T(1)$$
so that if $T(1) = 1$,
$$T(2^n) = 2^{n-1} + \sum_{k=1}^n k \cdot 2^k + \sum_{k=1}^{n-2} (2^{n-k-1}-1)2^k.$$
