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This question already has an answer here:

I am having troubles with a proof question.

Prove that for any $n\ge1$, $\sum_{i=1}^n f_i^2=f_nf_{n+1}$, where $f_n$ is the $n$'th Fibonacci number.

I have the base case and the induction hypothesis, and I know what I need to prove (substitute $n+1$ in for $n$'s on both sides of the equation) If someone can just guide me in the right direction on where to go, using induction that would be helpful. Thank You

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marked as duplicate by AlexR, Brian Rushton, Martin Sleziak, drhab, Lord_Farin Feb 4 '15 at 18:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Add $f_{n+1}^2$ to $f_nf_{n+1}$ and do a little manipulation. $\endgroup$ – André Nicolas Feb 4 '15 at 16:55
  • $\begingroup$ This is bound to be a duplicate. Searching... $\endgroup$ – AlexR Feb 4 '15 at 16:57
  • $\begingroup$ @james If you want to get more detailed help, you need to explain in more detail what you are having trouble with. Is your problem that you don't understand the idea of induction? Or are you having trouble with the $\sum$ notation? Or is it something else? As it is now, your question is vague enough that it's hard for anyone to help you without simply doing the whole problem, which might put you in violation of your school's ethics rules. $\endgroup$ – MJD Feb 4 '15 at 17:12
  • $\begingroup$ Write down your inductive assumption (in the above post), and write down what your inductive conclusion is. $\endgroup$ – DanielV Feb 4 '15 at 17:17
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Try adding $f_{n+1}^2$ to both sides of the equation you have, and using the properties of the fibonacci numbers to simplify the right side. Let me know if you need any help.

More explicitly, if $\Sigma_{i=1}^n f_i^2=f_n f_{n+1}$, then

$f_{n+1}^2+\Sigma_{i=1}^n f_i^2=f_n f_{n+1}+f_{n+1}f_{n+1}=f_{n+1}(f_n+f_{n+1})$ See where that can get you.

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  • $\begingroup$ Hello i'm really new to this type of mathematics, may you please elaborate on that $\endgroup$ – James Feb 4 '15 at 17:02
  • $\begingroup$ Thank you @Brian Rushton that helps a lot! $\endgroup$ – James Feb 4 '15 at 17:21

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