Prove by induction. How do I prove $\sum\limits_{i=0}^n2^i=2^{n+1}-1$ with induction? $\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.
I don't understand induction so I could use some help.
 A: The base case is the calculation $2^0=2-1$. Assume $\sum_{i=0} ^{n-1} 2^i=2^n-1$ (this is the induction step). Adding $2^n$ to both sides gives:
$$\sum_{i=0} ^{n} 2^i=2^n-1+2^n=2\cdot2^n-1=2^{n+1}-1,$$
closing the induction and finishing the proof. 
A combinatorial approach to the problem is as follows. The total number of subsets of a set of size $n$ is $2^{n}$. Let $N_i=\{1,2,...,i\}$. Then the total number of subsets of $N_{j-1}$ is $2^{j-1}$. Every element of each subset of $N_{j-1}$ is less than $j$, therefore affixing $j$ to each of these subsets creates an exhaustive list of the subsets of $N$ which have $j$ as their greatest element. 
As before, the total number of nonempty subsets of $N=\{1,2,...,n+1\}$ is $2^{n+1}-1$, since we are taking away the empty set. Another way to count this is to count the number of subsets of $N$ with $j$ as their greatest element. We can partition $P(N)\setminus \emptyset$, where $P(X)$ denotes the power set of $X$, as follows: let $X_i$ denote the set of subsets of $N$ which have $i$ as there greatest element. Then $P(N)\setminus \emptyset=\bigcup_{i=1} ^{n+1} X_i$, which implies:
$$\sum_{i=0} ^{n} 2^i=2^{n+1}-1$$
A: First let us check the base case i.e. $n=0$.
$$\sum_{i=0}^02^i=2^0=1$$
$$2^{0+1}-1=1$$
Therefore the statement is true for $n=0$. 

Now assume that the statement is true for $n$: $\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$. We must show that it is also true for $n+1$: 
$$\sum_{i=0}^{n+1}2^i=\sum_{i=0}^{n}2^i+2^{n+1}=2^{n+1}-1+2^{n+1}=2\cdot2^{n+1}-1=2^{n+2}-1 \ \Box $$

So (i) the statement is true for $n=0$; and (ii) if the statement is true for $n$, then it is also true for $n+1$. Hence, by induction, it is true for all $n \ge 0$.
A: 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 $n \geq 0$.
