Thread: Recursion/induction

Let a0 and a1 be distinct real numbers

Set an=(an-1+an-2)/2 for each n>=2

Use induction to show that:

an+1-an=(-1/2)^n*(a1-ao)

-----------

If we take fore n=2...a2=a1+a0/2

And then a3=(((a1+a0)/2)+(a1))/2

So a3-a2 should be equal to that WALBOA

The problem I'm having is how to show that P(n)--> P(n+1)

You can assume this is true for n, but I don't know how to show it true for n+1

Originally Posted by zhupolongjoe

Let a0 and a1 be distinct real numbers

Set an=(an-1+an-2)/2 for each n>=2

Use induction to show that:

an+1-an=(-1/2)^n*(a1-ao)

-----------

If we take fore n=2...a2=a1+a0/2

And then a3=(((a1+a0)/2)+(a1))/2

So a3-a2 should be equal to that WALBOA

The problem I'm having is how to show that P(n)--> P(n+1)

You can assume this is true for n, but I don't know how to show it true for n+1

If

then

Forgive me for my ignorance, but why do you use

an+1-an=(-1/2)^n?

The problem statement has a(n+1)-a(n)=(-1/2)^n*(a1-a0)

Does (a1-a0) equal 1 here? How would we know that?

Also, aren't you supposed to use the fact that a(n)=(a(n-1)+a(n-2))/2 in some way?

THank you.

Originally Posted by zhupolongjoe

Forgive me for my ignorance, but why do you use

an+1-an=(-1/2)^n?

The problem statement has a(n+1)-a(n)=(-1/2)^n*(a1-a0)

Does (a1-a0) equal 1 here? How would we know that?

Also, aren't you supposed to use the fact that a(n)=(a(n-1)+a(n-2))/2 in some way?

THank you.

I inserted the missing piece - thanks. I have used . It's in the third line. Replace with so .

ok, got it, thanks