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JRHarrison90
Group Title
Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.
 one year ago
 one year ago
JRHarrison90 Group Title
Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.
 one year ago
 one year ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
first 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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now 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 \(k+1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
that 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 \[(k+1)^3+2(k+1)\]
 one year ago

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

electrokid Group TitleBest ResponseYou've already chosen the best response.1
from 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"
 one year ago

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

JRHarrison90 Group TitleBest ResponseYou've already chosen the best response.0
Makes sense.
 one year ago
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