Prove that n^3+2n is divisible by 3.
We're going to try this with LaTeX and hope for the best.
Case 1: n = 1
$\displaystyle 1^3 + 2 \cdot 1 = 1 + 2 = 3$
Check!
Now assume the theorem is true for n = k. We need to show it is true for n = k + 1. So we assume
$\displaystyle k^3 + 2k = 3a$
where a is an integer.
Thus we need to show the same for
$\displaystyle (k + 1)^3 + 2(k + 1)$
$\displaystyle = k^3 + 3k^2 + 3k + 1 + 2k + 2$
$\displaystyle = (k^3 + 2k) + (3k^2 + 3k + 3)$
The first term is divisible by 3 due to our assumption at n = k. The second term is obviously divisible by 3.
etc, etc.
-Dan