How would I convert a recursive geometric sequence summation to a closed formula? How would I convert the below geometric (I assume based on the terms) recursive sequence summation to a closed formula?
$$a_1 = 1,\ \quad a_k = \sum_{i=1}^{k-1} a_i \ \quad for\ k \geqq 2$$
I've tried:
$$a_k = 2\frac{k-1}{k}$$
$$a_k = 2\frac{1-k^2}{1-k}$$
$$a_k = k\frac{1-k^2}{1-k}$$
But nothing seems to work correctly with the terms (with $a_1$ to $a_7$ being 1, 1, 2, 4, 8, 16, and 32 respectively). I'm pretty stuck and really not sure how to proceed so would appreciate any help.
UPDATE: Thanks for the help everyone, in truth it was a combination of the answers that helped me better understand how to proceed with this so for future users looking to understand this question better, I would advise going through all the answers and not just the chosen one.

We have $a_k = S_{k-1}$ then $a_k - a_{k-1} = S_{k-1}-S_{k-2} = a_{k-1}$ then for $k \ge 2$ we have the equivalent recurrence
$$
a_k = 2 a_{k-1}\Rightarrow a_k = c_0 2^k
$$
