anonymous
  • anonymous
using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
first of all maybe we can see why it is true without induction
anonymous
  • anonymous
or just do it by induction either way have you ever done a proof by induction?
anonymous
  • anonymous
i'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?

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anonymous
  • anonymous
no, base case is \(n=1\)
anonymous
  • anonymous
in which case you get 3, so done with the base case
anonymous
  • anonymous
next 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
  • anonymous
most 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
  • anonymous
one sec let me try to absorb what you said. thank you for helping me. i will reply in a min with questions
anonymous
  • anonymous
take your time, do the algebra
anonymous
  • anonymous
what is the significance of substituting n for k?
anonymous
  • anonymous
you substitute \(k\) for \(n\) to say what you get to assume is true
anonymous
  • anonymous
not \(n\) for \(k\)
anonymous
  • anonymous
ok, so by saying k we are just saying we assume it's true. this signifies induction?
anonymous
  • anonymous
and if k+1 is true then the proof is complete?
anonymous
  • anonymous
yes 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
  • anonymous
some 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
  • anonymous
are you working on the algebra?
anonymous
  • anonymous
i am trying to work on the algebra. this is what i have to solve: (k+1)^3+2(k+1)=3
anonymous
  • anonymous
oh no not \(=3\) no one says it is equal to 3
anonymous
  • anonymous
just 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
  • anonymous
oops i meant \[(k+1)^3+2(k+1)\]
anonymous
  • anonymous
go ahead and expand it, let me know what you get
anonymous
  • anonymous
still doing the algebra?
anonymous
  • anonymous
i 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
  • anonymous
multiply it all out there is no way around it
anonymous
  • anonymous
do you know how to cube \(k+1\) easily?
anonymous
  • anonymous
in general \[(a+b)^3=a^2+3a^2b+3ab^2+b^2\] in your case \(a=k,, b=1\)
anonymous
  • anonymous
and of course \(2(k+1)=2k+2\)
anonymous
  • anonymous
okay. i have to do (k+1)(k+1)(k+1) and multiply all of it.
anonymous
  • anonymous
yes, but i showed you how to do it in one step
anonymous
  • anonymous
yeah i got it. sorry about that
anonymous
  • anonymous
take your time, let me know what you get when you do all this algebra then we will be done in two steps after
anonymous
  • anonymous
what 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
  • anonymous
ok you should have \[k^3+3k^2+3k+1+2k+2\]
anonymous
  • anonymous
you forgot to distribute with \(2(k+1)\) i think
anonymous
  • anonymous
and also you wrote \(1^2\) but i am sure you didn't mean to
anonymous
  • anonymous
now only two steps separate out the \(k^3+2k\) part, that we get to assume is divisible by 3
anonymous
  • anonymous
you have \[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\]
anonymous
  • anonymous
the 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
  • anonymous
if the red wasn't divisible would it not be a proof?
anonymous
  • anonymous
yeah 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
  • anonymous
\[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\] \[\overbrace{\color{blue}{k^3+2k}}^{\text{ div by 3 by induction}}+\overbrace{\color{red}{3(k^2+k+1)}}^{\text{ has a factor of 3}}\]
anonymous
  • anonymous
and now we are done hope this helped clarify how to do a proof by induction they all pretty much work this way
anonymous
  • anonymous
i can't believe i found someone to walk me through that
anonymous
  • anonymous
yeah 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
  • anonymous
if i can find out how to donate to you i will :)

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