Here's the question you clicked on:
JRHarrison90
Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.
wow induction for this?
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
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 \(k+1\)
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 \[(k+1)^3+2(k+1)\]
but i get the feeling i am talking to myself, so i will be quiet now
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"
proof for \[2k^2+1=2\left({3a\over2}-2\right)+1=3a\]