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How do I prove this by induction?

Prove that for every natural number n, $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$

Here is my attempt.

Base Case: let $ n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true.

Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$

Our hypothesis is: $2^n = 2^{n+1} -1$

Here is where I'm getting off track. Lets look at the right side of the last equation: $2^{n+1} -1$ I can rewrite this as the following.

$2^1(2^n) - 1$ But, from our hypothesis $2^n = 2^{n+1} - 1$ Thus:

$2^1(2^{n+1} -1) -1$ This is where I get lost. Because when I distribute through I get this.

$2^{n+2} -2 -1$ This is wrong is it not? Am I not applying the rules of exponents correctly here? I have the solution so I know what I'm doing is wrong. Here is the correct proof. enter image description here

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    $\begingroup$ Your induction hypothesis and what you are trying to prove for induction are both incorrect. What you are trying to prove is that the sum of the powers of $2$ up to $n$ is equal to $2^{n+1}-1$. So your inductive hypothesis should be that this result is true for $k$; that is, that $$2^0+2^1+\cdots+2^k = 2^{k+1}-1.$$What you want to prove is that from this it follows that the result is true for $k+1$, that is, that$$2^0+2^1+\cdots+2^{k}+2^{k+1} = 2^{(k+1)+1}-1\quad\mbox{holds.}$$Instead, you are trying to prove that $2^m = 2^{m+1}-1$ for all $m$, which is false. $\endgroup$ Commented Feb 18, 2011 at 1:19

6 Answers 6

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An easy way to do this is using binary. Here's an idea of what I mean:

  • $2^0$ in binary is $1$
  • $2^1$ in binary is $10$
  • $2^2$ in binary is $100$

For a general rule:

$2^n$ in binary is $100\cdots0$ (n zeros)

Add those together and you get $2^0 + 2^1 + ... + 2^n$ in binary is $11...11$ ($n+1$ ones).

Now it's obvious that adding 1 to that gives you $$100\cdots00 \quad\text {($n+1$ zeros)}$$ Which as we all know is $2^{n+1}$.

Thus $2^{n+1} - 1$ is equal so the sum of powers of two up to $2^n$.

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    $\begingroup$ Thats "the simplest solution" to this problem (and the most elegant too, to me). However, OP's goal is to solve it by induction. $\endgroup$
    – JnxF
    Commented Jan 11, 2016 at 0:25
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Both

  • Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$
  • Our hypothesis is: $2^n = 2^{n+1} -1$

are wrong and should be

  • Inductive Step to prove is: $ 2^0 + 2^1 + ... + 2^n + 2^{n+1} = 2^{n+2} - 1$
  • Our hypothesis is: $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$

Add $2^{n+1}$ to both sides of the hypothesis and you have the step to prove since $2^{n+1}-1 +2^{n+1} = 2^{n+2} - 1$

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HINT $\ $ Here's the inductive proof for summing a general geometric series.

THEOREM $\rm\quad 1 + x + \cdots + x^{n-1}\ =\ \dfrac{x^{n}-1}{x-1}$

Proof $\ $ Base case: It is true for $\rm\ n = 1\:,\:$ viz. $\rm\ 1 = (x-1)/(x-1)\:$.

Inductive step: Suppose it is true for $\rm\ n = k\:.\ $ Then we have

$$\rm\ x^k + (x^{k-1} +\: \cdots\: + 1)\:\ =\:\ x^k +\frac{x^k-1}{x-1}\ =\ \frac{x^{k+1}-1}{x-1}$$

which implies it is true for $\rm\: n = k+1\:,\:$ thus completing the inductive proof.

The proof you seek is just the special case $\rm\ x = 2\ $.

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let $S = 2^0 + 2^1 + 2^2 + .... + 2^{n-1}$

so $2S = 2^1 + 2^2 + 2^3 + 2^n$

then $2S - S = S = (2^1 - 2^0) + (2^2 - 2^1) + ... + (2^{n-1} - 2^{n-1}) + 2^n = 2^n - 1$

and we got $2^0 + 2^1 + 2^2 + .... + 2^{n-1} = 2^n - 1$

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  • $\begingroup$ In the row 2S - S, the last bracket should be (2^n- - 2^{n-1}). But thank you so much for this elegant proof. I guess proof is in the eye of the reader after all. $\endgroup$
    – bytrangle
    Commented Jan 9 at 14:26
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I don't see the answer I like here, so I'm writing my own.

Basic proof:

We wish to prove $2^0 + 2^1 + ... + 2^{n-1} = 2^n - 1$ for all $n$. We can verify by inspection this is true for n=1. Next, assume that $2^0 + 2^1 + ... + 2^{n} = 2^{n+1} - 1$.

$(2^0 + 2^1 + ... + 2^n) + 2^{n+1} = (2^{n+1} - 1) + 2^{n+1} = 2 \cdot 2^{n+1} - 1 = 2^{n+2} - 1$, so we have shown $2^0 + 2^1 + ... + 2^{n-1} = 2^{n} - 1$ is true for all n.

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  • $\begingroup$ 2^n+1 - 1 will give you the correct answer, if we take n=1 then 2^1+1 -1 will come instead of 2^1 -1. So you will get 2^2-1 = 3. i.e. n=1 will give you 3==3, so the hypothesis is not wrong $\endgroup$
    – Prasanna
    Commented Jan 17, 2021 at 3:57
  • $\begingroup$ @PrasannaSasne Good point, I've updated my answer. See what you think. $\endgroup$ Commented Jan 18, 2021 at 7:55
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Proof is in the eye of the reader. A way I like to teach inductive proofs is to back up the inductive hypothesis by one, put the next item in it, then see if you match the claimed formula.

The base case is as described elsewhere: verify that the sum of $2^n$ is $2^{n+1}-1$ when $n=0.$ Tryng this out gives $1$ on the left and $1$ on the right, so the base case is legit.

The inductive hypothesis states that the sum of $2^i$ for $i=0, 1, 2, ...$ is $2^{n+1}-1$.

If that hypothesis is true at $n$, then it is also true at $n-1$. That is, it is assumed to be true, for the moment, at any value including $n-1$. If we believe this, then the sum up to $n-1$ is $2^{n-1+1}-1$.

Let's take that assumption and see what happens when we put the next item into it, that is, when we add $2^n$ into this assumed sum: $$2^{n-1+1}-1 + 2^n$$ $$= 2^{n} - 1 + 2^n$$by resolving the exponent in the left term, giving $$= 2\cdot2^n - 1$$ because there are two $2^n$ terms. Finishing up, $$= 2^1\cdot2^n - 1$$ $$= 2^{n+1} - 1$$by properties of exponents, giving us our claimed formula.

By adding the next term to the $n-1$ result, we get the claimed hypothesis, and therefore this pattern holds for any choice of $n$. Therefore, the hypothesis is true, and the presumed formula is correct.

Induction is a case of proof by "conveyor belt." You have an endless sequence of items starting at zero and your goal is to prove - by picking some arbitrary $n-1$ and $n$ somewhere along the way - that this formula holds as the sequence continues on forever. You use the sequence to help you verify the formula.

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