How to find a formula for the expression $2^0+2^1+\cdots+2^n$?

Question

Is there a way to simply this equation and express it in terms of $n$?

$$2^0+2^1+2^2+2^3+\cdots+2^n$$

Answer 1

$$\sum\limits_{i=0}^{n}2^i = 1+2+2^2+2^3+\cdots + 2^n = 1 + 2\left(1+2+2^2+\cdots + 2^{n-1}\right)=1+2\sum\limits_{i=0}^{n-1}2^i\text{.}$$ Now subtract $\sum\limits_{i=0}^{n-1}2^i$ from both sides: $$\sum\limits_{i=0}^{n}2^i-\sum\limits_{i=0}^{n-1}2^i=2^n$$ and $$1+2\sum\limits_{i=0}^{n-1}2^i-\sum\limits_{i=0}^{n-1}2^i=1+\sum\limits_{i=0}^{n-1}2^i\text{.}$$ This gives $$2^n-1=\sum\limits_{i=0}^{n-1}2^i$$ or $$2^{n+1}-1=\sum\limits_{i=0}^{n}2^i\text{.}$$

Answer 2

Let $S = 2^{0}+ 2^{1}+\ldots+2^{n}.$ Then, $2S = 2^{1}+ 2^{2}+\ldots+2^{n+1}= (S-2^{0})+2^{n+1},$ so $S=2^{n+1}-1$.

Answer 3

Let $S = 2^0+2^1+2^2+2^3+\cdots+2^n$.

In base $2$, it's $S = 1111\dots 111_2$ written with $n+1$ ones.

So $S + 1 = 1000\dots000_2$ (with $n+1$ zeros) $ = 2^{n+1}$.

Hence $S = 2^{n+1}-1$.