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Prove that f0 – f1 + f2 - … - f2n-1 + f2n = f2n-1 – 1. For n is all positive numbers. I have an idea to what I must do, but I can't figure what the base case is. I think it is f(0) = 0 and f(1) = 1. But then idk how to prove these?

Darneezie
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  • We can always use: http://math.stackexchange.com/questions/65011/prove-this-formula-for-the-fibonacci-sequence or http://www.proofwiki.org/wiki/Euler-Binet_Formula – lab bhattacharjee Mar 11 '14 at 03:58

2 Answers2

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Using Induction,

let $\displaystyle A(n):F_0-F_1+F_2-\cdots-F_{2n-1}+F_{2n}=F_{2n-1}-1$ holds true for $n=m$

$\displaystyle\implies A(m):F_0-F_1+F_2-\cdots-F_{2m-1}+F_{2m}=F_{2m-1}-1$

$\displaystyle\implies A(m+1):F_0-F_1+F_2-\cdots-F_{2m-1}+F_{2m}-F_{2m+1}+F_{2(m+1)}$

$\displaystyle=F_{2m-1}-1-F_{2m+1}+F_{2(m+1)}$

$\displaystyle=F_{2m-1}-1-F_{2m+1}+F_{2m+1}+F_{2m}$

$\displaystyle=F_{2m-1}-1+F_{2m}$

$\displaystyle=F_{2m+1}-1=F_{2(m+1)-1}-1$

Now establish the base case $n=1\implies A(1)$

lab bhattacharjee
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f(0) - f(1) + f(2) - f(3) + f(4) - f(5) + ..- - f(2n-1) + f(2n) = (f(2)+f(4)+..f(2n)) - (1 + f(3) + f(5) + .. + f(2n-1)) = (f(2n+1) - 1) - f(2n) = f(2n-1) - 1.

DeepSea
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