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.
 A: If you look at the sequence you have written out:




n
an




1
1


2
1


3
2


4
4


5
8


6
16


7
32




Looking at this table we guess that $a_k = 2^{k-2}$ for $k >= 2$
We want to prove this by induction.
the base case is simple $n=2$, as we have seen it holds above in the calculation.
For  the inductive step, assume it is true for all $i$ up to $a_n$, that is, assume $i \leq n \implies a_i = 2^{i-1}$. We want to show that it then must hold for $a_{n+1}$, or explicitly that $a_{n+1} = 2^{n-1}$
From the recursive relation, $a_{n+1} = \sum_{i=1}^{n} a_i$. Using our inductive assumption, we can rewrite this $\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} 2^{n-2}$.
As a partial sum of geometric series, $\sum_{i=1}^{n} 2^{n-2} = 2^{n-1}$.
But this shows that $a_{n+1} = 2^{n-1}$ and the inductive hypothesis is proven.
A: We claim that
$$a_k=2^{k-2}$$
for $n\geq 2$.
To prove that, we will use Strong Induction.
$$a_k = \sum_{i=1}^{k-1} a_i=1+\sum_{i=2}^{k-1}2^{i-2}$$
To calculate the sum, let's use $m=i-2$ and see that the sum becomes
$$1+\sum_{m=0}^{k-3}2^m=1+1\times \frac{2^{k-2}-1}{2-1}=1+2^{k-2}-1=2^{k-2}$$
This completes the proof.
A: 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
$$
a_k = 2 a_{k-1}\Rightarrow a_k = c_0 2^k
$$
