1) So I need to prove that
I did ;then I just don't know where to go.
Then There are 2 sequence ones that I don't even know where to begin
a) Let Fn be the n-th term of the Fibonacci Sequence 1, 1, 2, 3, 5, 8,.. Defined by the rule:
F(n+2) = F(n) + F(n+1) [ Fn is F sub n]
prove by induction that F1+F2 +....+Fn 2^(n+1)
b) The sequence a1, a2, a3, a4, ... an,... [again a sub n]
of positive integers defined by the rule:
a1 =1 a(n+1) = 3an (3 times a sub n) ( n = 1, 2,3,..) Prove that an = 3 ^(n-1) for n= 1,2,...
Hello again meg0529In fact, we can prove the strict inequality .
This is a little bit different from the usual induction proof. Notice that when we add the next term, , we would need to prove that the new sum < . So if we can prove that , we're there. We do this in, again, a slightly unusual way, by assuming that we have two consecutive values of for which it's true.
So let be the propositional function:
Then (Do you understand the notation I'm using here?)
is , which is true; is , which is also true. So is true .
(Do you understand this bit? It's a GP, first term 2, common ratio 2.)
Have you tried this question? If you can't to it, then you're really not understanding anything about induction at all. The here are simply the terms of a GP.b) The sequence a1, a2, a3, a4, ... an,... [again a sub n]
of positive integers defined by the rule:
a1 =1 a(n+1) = 3an (3 times a sub n) ( n = 1, 2,3,..) Prove that an = 3 ^(n-1) for n= 1,2,...
Just start with:
Let be the statement (propositional function) involving : , and then use the formula for to show that . Then check that is true, and you're done.
Show us your attempt if you'd like us to check it for you.
Grandad