Analysis

$$T(n) = \left\{ \begin{array}{ll} \Theta(1) & n = 1 \\ \Theta(n) + 2\sum_{k=1}^{n-1} T(k) & n > 1 \end{array} \right.$$

Want to show running time is at least exponential, so show T(n) = $\Omega(2^n)$.

By substitution method:

Show: T(n) = $\Omega(2^n)\geq c 2^n$

Assume: T(k) $\geq c 2^k$ for k < n

$$\begin{eqnarray*}T(n) & \geq & \Theta(n) + 2 \sum_{k=1}^{n-1} c2^k \\ & = & \T... ...^{n-1} -1) \\ & = & \Theta(n) + 2c2^n - 4c \\ & \geq & c 2^n \end{eqnarray*}$$

If $4c - \Theta(n) \leq 0$, or $c \leq \Theta(n)/4$ (okay for large enough n).

Thus, $T(n) = \Omega(2^n)$; still exponential.