TITLE: Prove that the sequence with $T(0)=1$ and $T(n) = 1 + \sum_{j=0}^{n-1}T(j)$ is given by $T(n)=2^n$
QUESTION [8 upvotes]: $T(0)=1 \\
T(n) = 1 + \sum_{j=0}^{n-1}T(j) \\
$
Show that $T(n) = 2^n$.
I know how to prove this by induction, but I would like to know how to show this using first principles.
Edit:
The way I want to solve this problem is manipulate $T(n)$ in such a way that it ends up as $2^n$.

REPLY [3 votes]: This is just the brute force method. It is obviously an overkill in this case, but I think it is worth to learn it. Put:
$$ f(x)=\sum_{j=0}^{+\infty}T(j)\, x^j. $$
The recurrence relation then gives $f(0)=1$ and:
$$\frac{f(x)}{1-x}=\sum_{j=0}^{+\infty}\left(\sum_{k\leq j}T(k)\right)x_j=\sum_{j=0}^{+\infty}(2T(j)-1)\,x^j = 2f(x)-\frac{1}{1-x},$$
hence:
$$f(x)=\frac{1}{1-2x}=\sum_{j=0}^{+\infty}2^j\,x^j,$$
from which:
$$T(j)=2^j$$
as wanted.