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Summation of series of product of Fibonacci numbers

Find $\sum_{i=1}^{N-1}F_{i + 1} F_{N + 4 - i}$

Is there a direct formula to calculate this, rather than actually sum all the terms?

EDIT: My initial summation was incorrect. Updated now.

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    I've changed your answer that should have been a comment, and it's comments, to comments on this question and Yury's answer.2012-09-09

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We seem to have a lot of "Sum of Products of Fibonacci Numbers" that are similar to each other, differing only in the limits of summation and the common total of the indices. To another question, I give a general answer that should cover all cases.

Applying $(5)$ from that answer to this question, yields $ \begin{align} \sum_{i=1}^{n-1}F_{i+1}F_{n+4-i} &=\sum_{i=2}^nF_iF_{n+5-i}\\ &=\frac{n-1}{5}(F_{n+4}+F_{n+6})+\frac15(F_{n-4}-F_{n+2}) \end{align} $

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I got the following formula $\sum_{i=1}^{N-1} F_i \cdot F_{N+4-i} = \frac{11}{5} (N-1)F_N+\frac{7}{5} N F_{N-1}$ by plugging in $F_n = \frac{1}{\sqrt{5}}\left( \left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)$ and simplifying the obtained expression.

(Update: I answered the original question of OP, not the new one.)

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    @AncientEgypt, here are the first values that my formula gives: 0, 5, 13, 31, 65, 130, 250, 469, 863, 1565, 2805, 4980, 8772, 15349, 26705, 46235, 79705, 136886, 234302, 399845, 680515, 1155385, 1957293, 3309096, 5584200, 9407525, 15823765, 26577559, 44579633, 74681770. All of them are integers. If you think that my formula is incorrect, please, let me know for what $N$ it gives an incorrect answer.2012-09-09