jmc

We calculate the first several terms directly and find the sequence starts $$1,1,2,4,8,16,\dots$$ It appears the $n$th term is $2^{n-2}$ for $n\ge 2$. Since $2^{12}=4096$, the first power of 2 that exceeds 5000 is $2^{13}=\boxed{8192}$.

Let's prove by induction that the $n$th term of the sequence is $2^{n-2}$ for all integers $n\ge 2$. The base case $n=2$ holds since the second term of the sequence is the sum of all the terms before it, which is just 1. For the induction step, let $n>2$ and suppose that the $(n-1)$st term is $2^{n-1-2}=2^{n-3}$. Then the sum of the first $n-2$ terms of the sequence is $2^{n-3}$, since the $(n-1)$st term is equal to the sum of the first $n-2$ terms. So the $n$th term, which is defined to be the sum of the first $n-1$ terms, is $$\underset{\text{sum of first }n-2\text{ terms}}{\underbrace{2^{n-3}}}+\underset{(n-1)\text{st term}}{\underbrace{2^{n-3}}}=2\cdot 2^{n-3}=2^{n-2}.$$ This completes the induction step, so the statement is proved for all $n\ge 2$.