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Suppose that $\{F_n\}$ is the sequence of Fibonacci numbers. There is a well-known result that $$\sum_{i=1}^nF_i^2 F_{i+1}=\frac{1}{2}F_nF_{n+1}F_{n+2}.$$ This is easy to prove by induction. I was wondering if there known results for the corresponding alternating sum, $$\sum_{i=1}^n (-1)^iF_i^2 F_{i+1}.$$ I've tried a few numerical hunts for a formula but couldn't find anything. Perhaps this problems has already been solved? Any hints or advice would be greatly appreciated! Ideal, would be an answer in terms of a factored product of Fibonacci and Lucas numbers, or something close!

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  • $\begingroup$ Using the formula for $F_n$ as a linear combination of geometric sequences, I think you should get an expression of your sum as a linear combination of geometric sums. $\endgroup$ – Joel Cohen Jul 11 '15 at 11:52
  • $\begingroup$ What happens if you use Binet? $\endgroup$ – J. M. isn't a mathematician Jul 11 '15 at 11:58
  • $\begingroup$ Thanks for responding, @Joel. Yes, that approach would give an answer. But I was wondering if there was something clean - a nice factorised expression, like the first quoted result. $\endgroup$ – Auslander Jul 11 '15 at 12:02
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If we use the explicit formula: $$ F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\overline{\sigma}^n\right),\qquad \sigma+\overline{\sigma}=1,\quad \sigma\overline{\sigma}=-1,$$ then we have:

$$ \sum_{i=1}^{n} (-1)^i F_i^2 F_{i+1} =\\= \frac{1}{5\sqrt{5}}\sum_{i=1}^{n}(-1)^i\left[\left(\sigma^{3i+1}-\overline{\sigma}^{3i+1}\right)-2(-1)^i\left(\sigma^{i+1}-\overline{\sigma}^{i+1}\right)-(-1)^{i+1}\left(\sigma^{i-1}-\overline{\sigma}^{i-1}\right)\right]=\\=\frac{1}{5}\sum_{i=1}^{n}\left[(-1)^i F_{3i+1}-2F_{i+1}+F_{i-1}\right]=\frac{-2(F_{n+3}-2)+(F_{n+1}-1)}{5}+\frac{1}{5}\sum_{i=1}^{n}(-1)^i F_{3i+1}$$

and we just need to compute the last sum: $$\begin{eqnarray*} \frac{1}{5\sqrt{5}}\sum_{i=1}^{n}(-1)^i\left(\sigma^{3i+1}-\overline{\sigma}^{3i+1}\right) &=&\frac{1}{5\sqrt5}\left[\frac{\sigma^2}{2}(-1+(-1)^n \sigma^{3n})-\frac{\overline{\sigma}^2}{2}(-1+(-1)^n \overline{\sigma}^{3n})\right]\\&=&\frac{1}{10}\left(-1+(-1)^n F_{3n+2}\right)\end{eqnarray*} $$ to have:

$$\sum_{i=1}^{n}(-1)^i F_i^2 F_{i+1} = \frac{5+(-1)^n F_{3n+2}-4 F_{n+3}+2 F_{n+1}}{10}.$$

As a curiosity, if we plug in $n=6$ we get $666$. $37$ is a divisor of $666$, but the first Fibonacci number that is divisible by $37$ is $F_{19}$, that is $F_{3n+1}$ in this case. So the chances that the alternating sum can be written as a nice product are very slim.

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  • $\begingroup$ This is a nice answer, for sure ! $\endgroup$ – Claude Leibovici Jul 11 '15 at 12:16
  • $\begingroup$ Thanks for your efforts @Jack. I will accept this answer if no one is aware of an approach that might yield a factored expression, like the one quoted in the question. $\endgroup$ – Auslander Jul 11 '15 at 12:16
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    $\begingroup$ @JackD'Aurizio There is an error in the calculation. For demonstration take $n=1$. The summation is $-1$. The answer reads $(-7 - F_{3} - 4 F_{4} + 2 F_{2})/10 = (-7 - 2 -12 + 2)/10 = -17/10$. For the case of $n=2$ the summation is 1, whereas the solution is $-15/10$. $\endgroup$ – Leucippus Jul 11 '15 at 13:58
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    $\begingroup$ @Leucippus: there must be some sign mistake, but the method is quite straightforward to follow, so feel free to fix it if you find the mistake before me. $\endgroup$ – Jack D'Aurizio Jul 11 '15 at 14:11
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    $\begingroup$ @Leucippus: it should be right now. $\endgroup$ – Jack D'Aurizio Jul 11 '15 at 14:32
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An alternative process is the following.

By using \begin{align} F_{n}^{2} \, F_{n+1} = \frac{1}{5} \left( F_{3n+1} - (-1)^{n} \, F_{n+2} \right) \end{align} then the desired series becomes \begin{align} S_{m} &= \sum_{n=1}^{m} (-1)^{n} \, F_{n}^{2} \, F_{n+1} \\ &= \frac{1}{5} \, \left[ \sum_{n=1}^{m} (-1)^{n} \, F_{3n+1} - \sum_{n=1}^{m} F_{n+2} \right] \\ &= \frac{1}{5} \left[ - \frac{1}{2} \left( F_{2} - (-1)^{m} F_{3m+2} \right) + F_{4} - F_{m+4} \right] \\ &= \frac{(-1)^{m} \, F_{3m+2} - 2 \, F_{m+4} + 5 }{10}. \end{align}

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