How to derive the closed form of the series

Question

The closed form is $2^r$ and the series is $1+2+4+8+\ldots$.

Answer 1

The closed form for the $n$th partial sum of the geometric series $1+2+4+8+...$ is $2^{n+1}-1$. This is the case since $$1 + 2 + 4 + \dots +2^n= 2\cdot(1 + 2 + 4+\dots + 2^n) - (1 + 2 + \dots + 2^n)= (2 + 4 + 8 + \dots + 2^{n+1}) - (1 + 2 + \dots + 2^n ) =2^{n+1} -1$$ as all other terms cancel out.

Answer 2

$a_n=2^n \text { for }n \ge 0$ $$S_n=\sum_{i=0}^n a_n$$ $$=1+2+...+2^n$$ $$2S_n-S_n=(2+2^2+...+2^{n+1})-(1+2+...+2^n)$$ $$(2-1)S_n=2^{n+1}-1$$ $$S_n=2^{n+1}-1 \text { for }n \ge 0.$$