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

1. anonymous

wow induction for this?

2. anonymous

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

3. anonymous

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 $$k+1$$

4. anonymous

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 $(k+1)^3+2(k+1)$

5. anonymous

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

6. anonymous

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"

7. anonymous

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

8. anonymous

Makes sense.