TITLE: Prove by induction. How do I prove $\sum\limits_{i=0}^n2^i=2^{n+1}-1$ with induction?
QUESTION [5 upvotes]: $\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.
I don't understand induction so I could use some help.

REPLY [4 votes]: The goal with mathematical induction is to prove the hypothesis true for the base case, and then show that if it holds for some arbitrary case (typically called $k$), it also holds for the next case ($k+1$).
For your problem, our base case occurs when $n = 0$.
$$\sum\limits_{i=0}^{0} 2^i = 2^0 = 1 = 2^1 - 1$$
Then, we assume that for some $k \geq 0$, the following holds:
$$\sum\limits_{i=0}^{k} 2^i = 2^{k+1} - 1$$
Now, if we can show that this being true implies that the hypothesis also holds for the $(k+1)$th case, the hypothesis must hold for all positive integers. So we have
$$\sum\limits_{i=0}^{k+1} 2^i = \sum\limits_{i=0}^{k} 2^i + 2^{k+1} = 2^{k+1} - 1 + 2^{k+1} = 2 \cdot 2^{k+1} - 1 = 2^{(k+1) + 1} - 1.$$
Thus, the hypothesis holds for all $n \geq 0$.