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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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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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. |
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