Induction and Recursion.. I have NO idea!

1MileCrash

I think I'm pretty good at standard induction. Never had a problem.

Induction and recursion is mercilessly whooping my ***.

1. The problem statement, all variables and given/known data

Let a₁ = 1. For each natural number n > 1, let a_n = 3a_n-1 - 1.

Prove that for each natural number n, a_n = 1/2(3^n-1 + 1)

2. Relevant equations



3. The attempt at a solution

WAT

I can "build it." And I don't even feel comfortable doing that. Let S be the set of n for which the theorem holds.

Let n = 2

Theorem holds.

Let n = 3

Theorem holds.

So 2,3 are elements of S.

(Do I have to show it for 2? Or more? Or just one? Why?)

Suppose that n >= 3 and {1,2,....,n} is a subset of S.

I have no idea what to do. At all. Can someone please explain recursion induction to me?

1MileCrash

I think I see..

I know that
a_n+1 = 3a_n - 1 because that's the definition.

And based on my assumption, a_n = 1/2(3[SUP]n-1[/SUP + 1)

So I can plug it in, and algebraically show that a_n+1 equals 1/2(3[SUP]n[/SUP + 1)

Am I on the right track?

micromass

Yes, that's the idea. Can you do that?

1MileCrash

Yes, that's what I tried and it worked out.

So for this example, I'd really only need to show it to be true for one value, right?

micromass

Yes, you only need to show it's true for n=1 and you need to show the induction step.

1MileCrash

Do you mean n = 2? If n = 1, a_n-1 is undefined?

micromass

Yes, check it for n=2 of course.

1MileCrash

Ok, got it. Thanks again!