Is it true that 3n – 1 is always an even number when n is a positive integer?
Hint
Test out the base case, break out the induction hypothesis, and make sure there aren't any stray counterexamples.
Answer
Nope; n = 2 is a counterexample: 3(2) – 1 = 6 – 1 = 5, which isn't even. There are plenty of others, too.
Example 2
Is it true that 2n + 1 is always an odd number when n is a positive integer?
Hint
You know the drill: test the base case, assume the induction hypothesis, and see if you can prove it.
Answer
Yes. The base case when n = 1 gives us 21 + 1 = 3, which is odd. We assume 2k + 1 is odd, then prove 2(k + 1) + 1 is odd, too. To do that, we rewrite the expression as 2(2k) + 1, then split it up even further to get 2k + 2k + 1. We know that first 2k is an even number 'cause it's a power of 2, and we know 2k + 1 is odd from our induction hypothesis, so (2k) + (2k + 1) = (even number) + (odd number). The sum of an even and odd number is always odd (like 4 + 3 = 7, etc.), so our sum must be an odd number. Boom, proof done.
Example 3
Is it true that 4n + 3n is always divisible by 2 when n is a positive integer?
Hint
It's pretty important to run through our induction steps in order. Start with the base case.
Answer
Bzzzt, that's false. We can't even get past our base case. When n = 1, we get 41 + 31 = 7. That's definitely not divisible by 2.