Hey is there any known combinatorial formula for nth fibonacci number?
(n+1)th fibonacci number is given by summation of r=0 to (round)n/2:C(n-r,r) Can someone verify the formula?Help!
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You’ll find proofs of your formula here and at the earlier questions linked from there. – Brian M. Scott Mar 20 '15 at 12:01
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See the first formula here http://en.wikipedia.org/wiki/Fibonacci_number#Use_in_mathematics and substitute $n\rightarrow n+1$ – gammatester Mar 20 '15 at 12:02
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Hint: Try to incorporate Binet's formula: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html
$$Fib(n) = \frac{\phi^n - \frac{(-1)^n}{\phi^n}}{\sqrt{5}}$$
Panglossian Oporopolist
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Right, just that I thought I should add a hint which may help you get an answer eventually... – Panglossian Oporopolist Mar 20 '15 at 12:03