Can anyone help me with this proof?
"Show that when is a positive integer."
Use induction and the recursion formula:
Let be the statement that , for all positive integer .
Base case: says that which is true.
Inductive step: Assume to be true, that is, for some positive . It remains to show that is also true, that is, .
........... What is the red equal to? Recall the recursive definition and everything should work out.
Alright, I've been trying to prove this through induction and using the Fibonacci Sequence.
when is a positive integer.
Assume this is true for , and prove this for by Induction Hypothesis
Fibonacci Sequence: for
Therefore, this is trivially true.
When using Fibonacci's Sequence we can use the relationships between . But I think we're missing a fourth term, so I added .
I tried to multiply the first terms to get the third and manipulate it by the Fibonacci's numbers and write one in terms of the other. But I ended up with some fractional expressions such as: and
I'm not sure how close I am to solving it or if there is an easier way. Could anyone finish this for me or work it out differently? Thanks.