1. ## 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

2. 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 $a_{n+1} - a_n = \left(- \frac{1}{2}\right)^n(a_1 - a_0)$

then

$
a_{n+2} - a_{n+1}
$

$
= \frac{1}{2} \left( a_{n+1} + a_n\right) - a_{n+1}
$

$
= - \frac{1}{2}\left( a_{n+1} - a_n \right)
$

$
= - \frac{1}{2} \left( - \frac{1}{2}\right)^n(a_1 - a_0)
$

$
= \left( -\frac{1}{2}\right)^{n+1}(a_1- a_0)
$

3. 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.

4. 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 $a_n = \frac{1}{2} \left(a_{n-1} + a_{n-2}\right)$. It's in the third line. Replace $n$ with $n+2$ so $a_{n+2} = \frac{1}{2} \left(a_{n+1} + a_{n}\right)$.

5. ok, got it, thanks