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anonymous
 5 years ago
Let \[a_{1}, a _{2}......\] be a Fibonacci series.
I have a book which has a problem under the "mathematical induction" chapter, which asks me to prove
\[a _{n+1}^{2}=a _{n}a _{n+2}=(1)^{n}\]
I think the statement is incorrect. For n=1, we don't have the statement correct, and I can't expect it to be correct for n>1 as \[(1)^{n}\] will always only have values either +1 or 1, and the expression \[a _{n}a _{n+2}\] will have increasing value.
Do you think I am correct, or is there really a prove for the statement.
Please provide me some help.
anonymous
 5 years ago
Let \[a_{1}, a _{2}......\] be a Fibonacci series. I have a book which has a problem under the "mathematical induction" chapter, which asks me to prove \[a _{n+1}^{2}=a _{n}a _{n+2}=(1)^{n}\] I think the statement is incorrect. For n=1, we don't have the statement correct, and I can't expect it to be correct for n>1 as \[(1)^{n}\] will always only have values either +1 or 1, and the expression \[a _{n}a _{n+2}\] will have increasing value. Do you think I am correct, or is there really a prove for the statement. Please provide me some help.

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+2}a_na_{n+1}^2=()^n\]is a true statement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the statement given there is not correct, am I right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your '=' sign is not a typo.?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When then it's incorrect.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should be a minus sign.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I told you a genius like you will need a look to solve the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't have any idea, what help you are providing just free of cost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyways, You are a busy person, so I must not hold you here, Bye

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) I haven't had a chance to look at the other question yet. I'm thinking the problem can be solved by proving first that linking the last contestant to fight the one left over at the beginning is the same as linking the new (k+1) guy to somewhere else in the system. Once that's done, the induction is as we said.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I should have more time in the middle of the week. When is STEP?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.020th June 2011 PM STEP Paper II takes place 22nd June 2011 PM STEP Paper III takes place 24th June 2011 AM STEP Paper I takes place

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, I have to go do some work. Like I said, middle of the week will be better.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I am taking several other similar exams, which you haven't heard of

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you intending to do with the results of the exams?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will email that to you. People will not like me saying such things here. They will think i am too presumptuous

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm off now. happy studies.
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