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 one year ago
Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.
 one year ago
Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.

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satellite73
 one year ago
Best ResponseYou've already chosen the best response.1wow induction for this?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1first check that it is true if \(n=1\) which it is because \(1^2+2\times 1=3\) and 3 is a factor of 3

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1now assume that it is true if \(n=k\) meaning assume that for all \(k\) you have \[k^3+2k\] is divisible by 3 now lets see if we can prove it is true for \(k+1\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1that is, see if you can show that \[(k+1)^3+2(k+1)\] is divisible by 3, if you get to assume that \(k^3+2k\) is mostly it is algebra to see if you can arrange to get a \(k^3+2k\) out of \[(k+1)^3+2(k+1)\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1but i get the feeling i am talking to myself, so i will be quiet now

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1from step2, \[k^3+2k=3a\\ k(k^2+2)=3a\] \[ (k+1)^3+2(k+1)=(k+1)[(k+1)^2+2]\\ \qquad=(k+1)[k^2+2k+1+2]\\ \qquad=(k+1)\left[{3a\over k}+2k+1\right]\\ \qquad=3a+2k^2+k+{3a\over k}+2k+1\\ \qquad=(3a)\left(1+{1\over k}\right)+3k+(2k^2+1) \] each of the three terms is a factor of "3"

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1proof for \[2k^2+1=2\left({3a\over2}2\right)+1=3a\]
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