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I know that I have to prove this with the induction formula. If proved the first condition i.e. $n=6$ which is divisible by $6$. But I got stuck on how to proceed with the second condition i.e $k$.

GNUSupporter 8964民主女神 地下教會
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Ella
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    Hint: note that $x=1$ is a root of $x^n-1$. – lulu Apr 26 '18 at 17:48
  • Yup I know that's how I have to solve it. But I got stuck at proving n=k and then n=k+1 – Ella Apr 26 '18 at 17:49
  • For the base case use $n=1$ rather than $n=6$ ... I assume you meant that for $n=1$ we get $7^n-1=6$ ... but you have to be clear about these things. Clarity and organization are key to these proofs! – Bram28 Apr 26 '18 at 17:49
  • Yup I did which is divisible by 6. But then the second condition is troubling me. – Ella Apr 26 '18 at 17:50
  • There is no need to use induction at all. $x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots +1)$. Now just let $x=7$. – lulu Apr 26 '18 at 17:51
  • @Ella OK, can you at least show what you did try? Can you state the inductive hypothesis? What is it that you can assume, and what is it that you need to show? Again, being clear on your assumptions and your goal is key here. – Bram28 Apr 26 '18 at 17:51
  • @Bram28 the question says that I need to prove the above equation with mathematical induction. I know the 2 condition need to be met for induction to be possible and this is the goal – Ella Apr 26 '18 at 17:54
  • @Ella Yes, I know, and I am trying to help you create exactly such an inductive proof. Now, you did the base case: good! But now, for the step, can you tell me (clearly! As in: a sentence) what you assume (this is the inductive hypothesis), and what you should try and show? And BTW: gotta run! Hopefully someone else can step you through this. – Bram28 Apr 26 '18 at 17:56
  • @Bram28 Yeah sure. Now I need to assume n is true for k. To prove the equation for $s(k)$ I need to input n=k+1 and then add the result of the term with the equation $s(k)$ – Ella Apr 26 '18 at 17:59
  • More duplicates: https://math.stackexchange.com/q/383379/290189, https://math.stackexchange.com/q/1184260/290189, etc – GNUSupporter 8964民主女神 地下教會 Apr 26 '18 at 18:41
  • @Ella Sorry to be so critical here, but you need to be much more careful and clear about your claims if you want to become successful in doing these kinds of proofs. Look back at what you are writing: "Assume $n$ is true for $k$" ... huh? $n$ and $k$ are both numbers, so you're saying something like "13 is true for 25"? That's not even a claim. Next sentence: "To prove the equation for s(k) .... with the equation s(k)" ...First of all: what is $s(k)$? Second, you say that $s(k)$ is an equation .. so you claim to have to "prove the equation for the equation"?! ... continued – Bram28 Apr 26 '18 at 22:22
  • @Ella "... input $n=k+1$" ... input this into what? Presumably something that has an $n$ .. but you don't mention anything like it. "... add the result of the term.." ... what term? ... and what result? What are you referring to that produces any kind of result? "... add the result of the term with the equation $s(k)$"... how do you add the result of something with an equation? ... Look, I hate to be harsh here, but it feels like you're using words and terms without any kind of understanding of what it all means. This is going to require a lot more help than what you can get on this site. – Bram28 Apr 26 '18 at 22:27
  • @Ella One final thought: If English is not your first language, I can explain some of your strange sentence constructions .. but certainly not all. Again, if you want to understand how induction works you're going to need more help than going to this site; there just isn't enough interactivity to step you through the whole process. You need a real live tutor ... is there someone who could indeed help you with these problems and how to think about them? – Bram28 Apr 26 '18 at 22:31
  • @Bram28 Thanks but I got this. My concepts weren't very clear. Yes english is my first language. I needed serious help on this topic but after one day of study I've got this. – Ella Apr 27 '18 at 15:38
  • @Ella Good to hear! :) – Bram28 Apr 27 '18 at 16:09
  • @Ella No, I don't. Sorry! And if you're looking for a tutor, I think you should try and find someone who you can meet with in person, rather than online; communication is just so much better in person! – Bram28 Apr 27 '18 at 16:21

5 Answers5

1

We have $$7\equiv 1\mod 6$$ then $$7^n\equiv 1^n=1\equiv 1\mod 6$$ so $$7^n-1\equiv 0 \mod 6$$

Dr. Sonnhard Graubner
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Hint: $7^{n+1}-1=7^{n+1}-7^n+7^n-1=6\times 7^n+7^n-1$.

José Carlos Santos
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Induction step:n+1.

$7^{n+1} -1 = 7\cdot 7^{n} -1=$

$ (6+1)(7^n) -1=$

$6 \cdot 7^n +(7^n -1).$

By hypothesis $(7^n-1)$ is divisible by $6$, hence the above sum is divisible by $6.$

Peter Szilas
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Let suppose that the result is true for $n=k$ i.e $7^k-1$ is divisible by $6$.

Then: $$7^{k+1}-1=7 \times (7^k-1)+7-1=7 \times (7^k-1)+6$$ but $(7^k-1)$ is divisible by $6$ so $7 \times (7^k-1)$ is also divisible by $6$ and finally $(7^k-1)+6$ is divisible by $6$.

Delta-u
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We have: $$7^n-1=(7-1)(7^{n-1}+7^{n-2}+\dots +7+1) = 6(7^{n-1}+7^{n-2}+\dots +7+1) $$

Which definitely is divisible by $6$.

Anastassis Kapetanakis
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