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)
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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
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.