To go a little farther than the comments, your condition is $$2a_n = \sum_{k=0}^{n-1} a_k$$ (unlike the comments, I am calling the first element of the sequence $a_0$). Note that $$\sum_{k=0}^{n-1} a_k = a_{n-1} + \sum_{k=0}^{n-2} a_k$$ and $$2a_{n-1} = \sum_{k=0}^{n-2} a_k$$ So $$2a_n = a_{n-1} + 2a_{n-1} = 3a_{n-1}\\\a_n =\left(3\over2\right)a_{n-1}$$
Hence your sequence is $$a_n = \left(3\over 2\right)^na_0$$