Recurrence Relation $T_n=\sum_{r=0}^{n-1} T_r+2^n$ From a recent solution I posted here, working from an alternative path would have led to the following recurrence relation which involves a summation term:
$$T_n=\sum_{r=0}^{n-1}T_r+2^n; \qquad T_0=1\qquad (n>1)$$
How can this be solved?
Edit 3 (replaces previous edits)
Initial condition is $T_0=1$. For the problem as stated, $T_1=3$ (not $4$) and does not need to be specified as an initial condition. 
This is actually different from the previous problem, which would have resulted in $T_n=\sum_{r=0}^{n-1}+2^{n+1}-1$. 

For clarification
For the previous problem that I referred to here, the recurrence relatioships, the first two terms, and the closed form solution are as follows:
$$T_n=2T_{n-1}+\color{blue}{2^n}\\
T_n=\sum_{r=0}^{n-1}T_r+\color{blue}{2^{n+1}-1}\\
T_0=1, T_1=\color{blue}4\\
T_n=(\color{blue}n+1)2^n$$
For the present problem as posted here, the recurrence relatioships, the first two terms, and the closed form solution are as follows:
$$T_n=2T_{n-1}+\color{red}{2^{n-1}}\\
T_n=\sum_{r=0}^{n-1}T_r+\color{red}{2^n}\\
T_0=1, T_1=\color{red}3\\
T_n=(\color{red}{\frac n2}+1)2^n$$

Use generating functions. Define $T(z) = \sum_{n \ge 0} T_n z^n$, adjust indices in the recurrence:
$\begin{align}
   T_{n + 1} = \sum_{0 \le r \le n} T_r + 2 \cdot 2^r
\end{align}$
Multiply by $z^n$ and sum over $n \ge 0$, recognize some sums in the result:
$\begin{align}
   \frac{T(z) - T_0}{z}
     = \frac{T(z)}{1 - z} + \frac{2}{1 - 2 z}
\end{align}$
Plug in $T_0 = 1$, solve for $T(z)$ as partial fractions:
$\begin{align}
   T(z)
     &= \frac{1 - z}{1 - 4 z + 4 z^2} \\
     &= \frac{1 - z}{(1 - 2 z)^2} \\
     &= \frac{1}{2 (1 - 2 z)^2} + \frac{1}{2 (1 - 2 z)}
\end{align}$
Using the binomial theorem and a geometric series:
$\begin{align}
   T_n
     &= \frac{1}{2} \cdot (-1)^n \binom{-2}{n} \cdot 2^n
          + \frac{1}{2} \cdot 2^n \\
     &= \frac{1}{2} \binom{n + 2 - 1}{2 - 1} \cdot 2^n
          + \frac{1}{2} \cdot 2^n \\
     &= (n + 2) \cdot 2^{n - 1}
\end{align}$
