# Fibonacci Sequence and odd/even addition [duplicate]

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?

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## marked as duplicate by lab bhattacharjee, Claude Leibovici, Stefan Hansen, Sami Ben Romdhane, Michael HoppeMar 11 '14 at 8:34

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 $n=1\implies A(1)$

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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.

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