# Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer

Letting $f_n$ be the Fibonacci numbers, show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer.

Just some homework help. Need to prove. Thank you in advance.

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What is $f_n$? Fibonacci I assume.... –  N. S. Nov 14 '11 at 21:32
Then it is not true ;) The problem should specify what $f_n$ is, and if not, it was probably defined earlier in another problem or textbook... –  N. S. Nov 14 '11 at 21:36
Think about the problem the following way: pick any sequence which verifies this relation, pick any $n$ you want and then change ONLY $f_{2n}$... What happens to the equality? So, unless you know exactly what $f_n$ is for all $n$, you can't prove this is right... –  N. S. Nov 14 '11 at 21:41
Yes, it is Fibonacci. Doesn't claim on the assinment, but does reference a set of problems from the book. My mistake, I apologize. –  Chris Nov 14 '11 at 22:07
Yet another case where homework difficulties would better be solved by asking the instructor, rather than here! –  GEdgar Nov 14 '11 at 22:16

Hint 1: Induction...

Hint 2: replace $f_n$ by their "closed" form, and see how you can calculate those sums.. It is easy...

Each hint leads to a different solution...

P.S. I assumed that $f_n$ is the Fibonacci sequence, I am pretty sure it is...

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Thank you for the prompt answer, by the way. –  Chris Nov 14 '11 at 22:56
In case you already now the formula for the sum of first $n$ Fibonacci numbers $$\sum_{j=0}^n F_j = F_{n+2} - 1$$ you can use it as follows:
$F_0-F_1+F_2-F_3+\dots-F_{2n-1}+F_{2n}=$ $F_0+(F_2-F_1)+(F_4-F_3)+\dots+(F_{2n}-F_{2n-1})=$ $0+F_0+F_2+\dots+F_{2n-2}=$ $0+(F_0+F_1)+\dots+(F_{2n-4}+F_{2n-3})=$ $\sum_{k=0}^{2n-3} F_n = F_{2n-1}-1.$
A proof of the above formula for the sum of the first $n$ Fibonacci numbers can be found e.g. at proofwiki.