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anonymous
 one year ago
using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.
anonymous
 one year ago
using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first of all maybe we can see why it is true without induction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or just do it by induction either way have you ever done a proof by induction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm supposed to solve it using induction. it's obvious that it's true. even if i plug in 0 it works. but, i don't understand understand how to set this up. i know i need a base case. would that be n=0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, base case is \(n=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in which case you get 3, so done with the base case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0next you assume it is true if \(n=k\) in other words you say "we assume \(k^3+2k\) is divisible by \(3\)" and using that, prove it is true if \(n=k+1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0most of it now is algebra from here on in replace \(n\) by \(k+1\) and separate out a part of the result that looks like \(k^3+2k\) and the rest will be obviously divisible by \(3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one sec let me try to absorb what you said. thank you for helping me. i will reply in a min with questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take your time, do the algebra

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the significance of substituting n for k?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you substitute \(k\) for \(n\) to say what you get to assume is true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so by saying k we are just saying we assume it's true. this signifies induction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if k+1 is true then the proof is complete?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that is the "induction hypothesis" assume it is true if \(n=k\) then you use that to prove it is true if \(n=k+1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0some people cheat and assume it is true for \(n\) then show it is true for \(n+1\) but really we should change the variable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you working on the algebra?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am trying to work on the algebra. this is what i have to solve: (k+1)^3+2(k+1)=3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no not \(=3\) no one says it is equal to 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just take \[(k+1)^2+2(k+1)\] and expand it, then take out the part that looks like \(k^3+2k\) and separate it from the rest

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops i meant \[(k+1)^3+2(k+1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0go ahead and expand it, let me know what you get

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0still doing the algebra?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am.. i feel retarded. thanks again for helping :\ i am trying to figure out what else i do after (k+1)((k+1)^2)+2(k+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply it all out there is no way around it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know how to cube \(k+1\) easily?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in general \[(a+b)^3=a^2+3a^2b+3ab^2+b^2\] in your case \(a=k,, b=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and of course \(2(k+1)=2k+2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay. i have to do (k+1)(k+1)(k+1) and multiply all of it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, but i showed you how to do it in one step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i got it. sorry about that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take your time, let me know what you get when you do all this algebra then we will be done in two steps after

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what did you mean by separate it from the rest? it's all multiplied out. k^2+3k^2(1)+3k(1)^2+1^2 + 2k+1 but there is no right side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok you should have \[k^3+3k^2+3k+1+2k+2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you forgot to distribute with \(2(k+1)\) i think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also you wrote \(1^2\) but i am sure you didn't mean to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now only two steps separate out the \(k^3+2k\) part, that we get to assume is divisible by 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have \[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the blue part is divisible by 3 "by induction" we got to assume that the red part is divisible by 3 because... well because it is you can factor a 3 out of it if you want to be formal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the red wasn't divisible would it not be a proof?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah you need that part to be divisible by 3 the blue part is the induction hypotheses the red part is what you have to show is divisible by 3, but there is not much to show since each term has a 3 in it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\] \[\overbrace{\color{blue}{k^3+2k}}^{\text{ div by 3 by induction}}+\overbrace{\color{red}{3(k^2+k+1)}}^{\text{ has a factor of 3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and now we are done hope this helped clarify how to do a proof by induction they all pretty much work this way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can't believe i found someone to walk me through that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it is usually a thankless task because most people have no idea about what it means so you have to explain every step of the way, and they get frustrated really it is something you have to see done a few times to understand the logic glad it helped

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if i can find out how to donate to you i will :)
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