Question

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

Answer 1

wow induction for this?

Answer 2

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

Answer 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 \(k+1\)

Answer 4

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
\[(k+1)^3+2(k+1)\]

Answer 5

but i get the feeling i am talking to myself, so i will be quiet now

Answer 6

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"

Answer 7

proof for \[2k^2+1=2\left({3a\over2}-2\right)+1=3a\]

Answer 8

Makes sense.