anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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DebbieG
  • DebbieG
Oh I love this. OK, here's what I THINK you do. First, show it for n=1. That's pretty trivial (just sum the cubes of 1, 2 and 3). You get a multiple of 9. So it holds for n=1. Now assume it holds for n. Then for n+1 you have: \((n+1)^3+(n+2)^3+(n+3)^3\) as the sum of the cubes of the 3 consecutive numbers, right? Expand that. (I cheated and used Wolfram, lol, but it can certainly be done by using the formula for the cube of a binomial). Look at what you end up with .... you will have an expression that you can re-write as: n(....something that must be a natural number) + 9*(a natural number) Lets write this as: n*a+9*b Since n is divisible by 9, we know that n=9k for some k. SO.... can you finish it from here? :)
DebbieG
  • DebbieG
And for that: n(....something that must be a natural number) + 9*(a natural number) ... be sure you don't gloss over why that stuff in the ( )'s "must be a natural number". It's obvious that is must, but it's an important part of the proof to state so and why. :)

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DebbieG
  • DebbieG
Geez lots of lurkers... I see you all! :P lol
ybarrap
  • ybarrap
We use two tricks to prove this statement. One, is that if two numbers are multiples of 9, then so are their differences. Second, a difference of cubes has a nice formula that we utilize with the 1st trick: \(a^3-b^3=(a-b)(a^2+ab+b^2)\): $$ \text{Proof that: }f(k)=k^3+(k+1)^3+(k+2)^3=9n\text{, where }k,n\in\Bbb N\\ \text{Basis:}\\ \quad f(0)=1^3+2^3=9\text{, which is a multiple of 9 with }n=1\\\\ \text{Induction Step:}\\\\ \quad\text{Assume }f(k)=9m\text{,where }k,m\in\Bbb N\\ \quad f(k+1)=9p\equiv f(k+1)-f(k)=9q\text{, where }k,p,q\in\Bbb N\\ \quad f(k+1)=(k+1)^3+(k+2)^3+(k+3)^3\\ \quad\text{So, }f(k+1)-f(k)\\ \quad=(k+3)^3-k^3\\ \quad=(k+3-k)((k+3)^2+k(k+3)+k^2)\\ \quad=3(k^2+6k+9+k^2+3k+k^2)\\ \quad=3(3k^2+9k+9)\\ \quad=3(3)(k^2+3k+1)\\ \quad=9q\text{, where }q\in\Bbb N\\\\ \text{Therefore, }f(k+1)=9p\text{, QED.} $$
DebbieG
  • DebbieG
This definitely looks slicker than mine...although I need to stare at it a little bit longer, but just one question. Don't you need to establish first for f(1), not for f(0)? 0 is not a natural number...?
ybarrap
  • ybarrap
I don't think slicker,all proofs are beautiful. I proved initially using binomial and was deflated once the difference of cubes struck me and so I had to share. Whether n = 0 or n = 1 depends on the definition of the natural numbers. If 0 is considered a natural number, as is common in the fields of combinatorics and mathematical logic, the base case is given by n = 0. If, on the other hand, 1 is taken as the first natural number, then the base case is given by n = 1.
DebbieG
  • DebbieG
Ah, gotcha. I never thought of 0 as a natural number but I see what you're saying. I looooooved doing proofs like this when I was a student, but that's been a while. :) It was fun to figure out how this one works! I agree, proofs are beautiful.... always so nice to see it all come together and write that "QED" (although my own quirky thing was to make the "Q" into a smiley face, LOL!!)

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