Prove that $\sum_{i=0}^{n-1} {2^i} = 2^n -1$

Question

I need to prove that $$\sum_{i=0}^{n-1} {2^i} = 2^n -1.$$ I tried induction but something didn't work.

Answer 1

Induction step:

$$\sum_{i=0}^n2^i=\sum_{i=0}^{n-1}2^i+2^n=2^n-1+2^n=2^{n+1}-1.$$

Answer 2

Here is a direct proof, using telescopic sums :

Let $ n $ be a positive integer, we have : \begin{aligned}\sum_{k=0}^{n-1}{2^{k}}=\sum_{k=0}^{n-1}{2^{k}\left(2-1\right)}=\sum_{k=0}^{n-1}{\left(2^{k+1}-2^{k}\right)}=2^{n}-1\end{aligned}

Answer 3

This can be proved by induction, test for $n=2$:

$$\sum_{i=0}^{2-1} 2^{i}=2^0+2^1=3=2^2-1$$

Now, suppose this is valid for some $k\geq 2$,

$$\sum_{i=0}^{k-1} 2^{i}=2^k-1$$

Now check if it is valid for $k + 1$

$$\sum_{i=0}^{k}2^i=2^k + \sum_{i=0}^{k-1}2^i=2^k+2^k-1=2^{k+1}-1$$