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Trying to prove that the sum of odd-index consecutive Fibonacci numbers is the next even-index Fibonacci number. I have a gap in my proof that I cannot figure out. I know that induction would be easier and I have already done it that way, I am looking to use alpha and beta. $$\sum_{i=1}^n F_{2i-1} = F_{2n}$$

$$\begin{align*} \sum_{i=1}^n F_{2i-1}&=\sum_{i=1}^1 \frac{1}{\sqrt{5}}(\alpha^{2i-1}-\beta^{2i-1})\\ &=\frac{1}{\sqrt{5}}\left(\sum_{i=1}^1\alpha^{2i-1}-\sum_{i=1}^1\beta^{2i-1}\right)\\ &\;\;\vdots\\ &=\frac{1}{\sqrt{5}}[(\alpha^n-\beta^n)(\alpha^n+\beta^n)]\\ &=\frac{1}{\sqrt{5}}(\alpha^{2n}-\beta^{2n})\\ &=F_{2n} \end{align*}$$

Caleb Stanford
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Brad
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    There's a straightforward proof by induction. Suppose that it's true for a given $n$ - i.e., that $F_{2n}=\sum_{i=1}^nF_{2i-1}$; then $\sum_{i=1}^{n+1}F_{2i-1}=F_{2n+1}+\sum_{i=1}^nF_{2i-1}=F_{2n+1}+F_{2n}$... – Steven Stadnicki Apr 24 '15 at 18:24
  • I dont want to use induction. I need to use alpha and beta. – Brad Apr 24 '15 at 18:25
  • duplicate but the answers here are better: https://math.stackexchange.com/questions/673218/prove-for-any-n-geq-1-recursion – gt6989b Feb 04 '21 at 17:37

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HINT: Induction is much easier, but if you really must use Binet’s formula, note that your summations are geometric, with ratios $\alpha^2$ and $\beta^2$, respectively; you can use the familiar formula for the sum of a finite geometric series. You also know that $\alpha$ and $\beta$ satisfy the equation $x^2-x-1=0$, so $\alpha^2-1=\alpha$ and $\beta^2-1=\beta$; this information will help you simplify the sums of the geometric progressions.

Brian M. Scott
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You can use (if $x^2-1\neq 0$): $$\sum_{i=1}^{n}x^{2i-1}=\frac{x^{2n+1}-x}{x^2-1} $$

Jean-Claude Arbaut
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Elaqqad
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