# How do I prove this by induction? (sum of powers of 2)

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.

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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. –  Arturo Magidin Feb 18 '11 at 1:19

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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