Math Is Fun Forum

darfmore

Member

Registered: 2012-03-02

Posts: 4

Induction Proof, Am I doing it right?

k

Last edited by darfmore (2012-03-10 11:07:09)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Induction Proof, Am I doing it right?

hi darfmore

Welcome to the forum!

The initial step is OK.

LATER EDIT: Having slept on it, I've realised there's a bit more to do here.

Because each term depends on the two previous terms it is not enough just to justify a1.  You need to also show the formula works for a2.

Then, a1 and a2 lead to a3.
a2 and a3 lead to a4.
a3 and a4 lead to a5
and so on.

The rest has all the right bits of algebra but in the wrong order.

For n + 1,
a_n+1 = 3^n+1 - (-2)^n+1 = a_n + 6a_n −1
3*3^n + 2(-2)^n = a_n + 6a_n-1

This looks back to front to me.  You should be starting with the definition

and substituting

plus

Then similar algebraic steps to what you have done should enable you to reach

which means you have done the inductive step.

Think of induction as being like climbing a ladder.

First you have to be at the bottom of a ladder.  That's the initial step.

Then, if you are on a step on the ladder, can you get to the next step?  That's the inductive step.

If you know how to climb from one step to the next and you are on the bottom step, then you can reach all the steps.

LATER EDIT:  In this problem you have to use what is called strong induction.  Each step up the ladder depends not just on being able to get from the previous step;  rather, you have to use earlier steps too.  To continue my ladder analogy, we have one foot on step n-1 and one foot on step n.  Taking these two steps together, we can push up to step n + 1.  In order to begin the climb we have to have a foot on a1 and a foot on a2.

Bob

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Children are not defined by school ...........The Fonz
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Sometimes I deliberately make mistakes, just to test you!  …………….Bob