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I just started learning proof by induction in class, but got a problem requiring proof by strong induction.

Here is the problem.

Prove by strong induction: $$\sum_{i=1}^n 2^i = 2^{n+1} - 2$$

I've done the base, showing that the statement holds for $n=1$, $n=2$, and $n=3$. (I won't show the simple math here). For $n=k$, the statement would be $2^{k+1}-2$. But that's where I get stuck, as I'm still trying to grasp the concept of strong induction.

For $n=k+1$, do I do the following and simplify?

$$\sum_{i=1}^{k+1} 2^i = \sum_{i=1}^k 2^i + 2^{(k+1)+1} - 2$$ $$=[2^{k+1}-2]+[2^{k+2}-2]$$ $$=\text{etc}\ldots?$$

Martin Sleziak
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Greg
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4 Answers4

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For strong induction, a proof goes something like this:

Proof of a base case: here $n= 1$ will do:

$2 = 4 - 2$.

The only thing different between "strong" and "regular" induction is how we state the inductive step:

We assume that for ALL $n_0 \leq k < n$, the theorem holds, and then use that to show it holds for $k = n$. This means we have access to ANY previous non-negative integer $k$ in that range, not just "the previous integer", $n-1$.

In this case, that means we can assume the result for all $1 \leq k < n$.

Of course, since $k = n - 1 < n$, we can use that case, too.

So we have:

$\sum\limits_{i = 1}^n 2^i = \sum\limits_{i = 1}^{n-1} 2^i + 2^n = (2^n - 2) + 2^n = \dots ?$

David Wheeler
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  • So the essential difference between regular and strong induction is that in regular induction, you assume that the arbitrary $k$ is $n<k$, whereas in strong induction, the relationship is $k<n$, giving access to any integer $i<k<n$? – Greg Aug 01 '15 at 03:17
  • It probably would have been worth mentioning here that strong induction is not needed at all--from OP's question, it seems he may think it actually is important to use strong induction, but that clearly isn't the case. – Daniel W. Farlow Aug 01 '15 at 04:08
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    In regular induction you go "step by step", using the $n-1$ case to prove the $n$ case. Thus if your base case is $n_0$, then $n_0$ proves $n_0 + 1$, and the $n_0 + 1$ case proves the $n_0 + 2$ case, etc. In strong induction, you are allowed to use any case between your base case and $n$. This is often helpful when you want to split $n$ into smaller factors, for example. The "$i$" in this problem, by the way, is just a "dummy variable" (used as an index of summation), and any other letter could be used. – David Wheeler Aug 01 '15 at 05:05
  • @Daniel I was able to prove the statement using regular induction, but the problem specifically says to prove using strong induction in order to nail the concept. If only life made sense :( – Greg Aug 01 '15 at 14:38
  • @David That makes mich better sense now. I tried using $n-2$ as well and it worked! Thanks so much. – Greg Aug 01 '15 at 14:41
  • @Greg Some people simply choose to always use strong induction and that's fine, but the point is that it is really not necessary in this case. If you are learning weak/strong induction for the first time, I would expect a better problem from a teacher to highlight when/why strong induction needs to be used, and this problem is not one of those times. Did you look at the weak vs. strong inducton link? One problem is proved with strong induction though only weak is needed. – Daniel W. Farlow Aug 01 '15 at 14:55
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Your calculation $$\sum_{i=1}^{k+1}2^i=\sum_{i=1}^{k}2^i+2^{(k+1)+1}-2$$ is false. What's true is that $$\sum_{i=1}^{k+1}2^i=\sum_{i=1}^k2^i+2^{k+1},$$ and $2^{k+1}\neq 2^{(k+1)+1}-2$ in general. You can apply the induction hypothesis to the first term on the right to get what you want, although this is normal induction, rather than strong induction.

neth
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  • Right, I understand how to prove the statement by regular induction, but it's the strong induction that I'm having trouble wrapping my mind around. – Greg Aug 01 '15 at 03:14
  • In that case, David addresses your question better. It's just that $k<k+1$, if if we can assume for all $<k+1$, we can certainly assume for $k$. – neth Aug 01 '15 at 03:17
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Suppose $\sum_{i=1}^k 2^i = 2^{k+1} - 2 $ for all $k < n$.

Then, choose some $k$ with $1 < k < n$.

$\begin{array}\\ \sum_{i=1}^n 2^i &=\sum_{i=1}^k 2^i +\sum_{i=k+1}^n 2^i\\ &=2^{k+1} - 2 +2^k\sum_{i=k+1}^n 2^{i-k}\\ &=2^{k+1} - 2 +2^k\sum_{i=1}^{n-k} 2^{i}\\ &=2^{k+1} - 2 +2^k(2^{n-k+1}-2)\\ &=2^{k+1} - 2 +2^{n+1}-2^{k+1}\\ &=2^{n+1} - 2 \\ \end{array} $

This is the only way I could think of to use strong induction.

marty cohen
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We have to prove, $$\sum_{i=1}^{n}2^{i}=2^{n+1}-2$$ Step 1: Setting $n=1$ in the above equality, we get $$\sum_{i=1}^{1}2^{i}=2^{1+1}-2\iff 2=4-2\iff 2=2$$ hence it holds for $n=1$

Step 2: Assuming that it holds for $n=k$, we get $$\sum_{i=1}^{k}2^{i}=2^{k+1}-2$$

Step 3: Now, setting $n=k+1$ in the given equality, we get $$\sum_{i=1}^{k+1}2^{i}=2^{k+1+1}-2 $$$$\sum_{i=1}^{k}2^{i}+2^{k+1}=2^{k+2}-2$$ $$\sum_{i=1}^{k}2^{i}=2.2^{k+1}-2-2^{k+1}$$ $$\sum_{i=1}^{k}2^{i}=2^{k+1}-2$$ Which is true from (2), hence it holds for $n=k+1$

It is clear from (1), (2) & (3) that the given equality holds for all positive integers $\color{blue}{n\geq 1}$

Harish Chandra Rajpoot
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