I have been given an assignment question that asks for a simple closed form of the following sequence:

$$G_n=\left|\begin{array}{cc} F_n & F_{n+1}\\ F_{n+1} & F_{n+2} \end{array}\right|$$

I have tried taking the determinant, but substituting in the closed form of the Fibonacci sequence leads to nothing simple at all.


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    $\begingroup$ Closed form will work. Or else compute the values for a few $n$, make a conjecture based on the results, and prove it by induction. $\endgroup$ – André Nicolas Sep 28 '15 at 5:04
  • $\begingroup$ By your expression do you mean the following? $$\left|\begin{array}{cc} F_n & F_{n+1}\\ F_{n+1} & F_{n+2} \end{array}\right|$$ $\endgroup$ – Samrat Mukhopadhyay Sep 28 '15 at 5:15
  • $\begingroup$ Yes, that is what I mean. Except the first entry is F_n and it is the determinant of the matrix. I do not know how to use Latex or the formatting on this site. $\endgroup$ – Edward Nashton Sep 28 '15 at 5:19
  • $\begingroup$ @EdwardNashton, $the \left|\right|$ sign itself denotes the determinant, you do not need to put a $det$ before it. $\endgroup$ – Samrat Mukhopadhyay Sep 28 '15 at 5:24
  • $\begingroup$ Oh, OK. Thanks. :) $\endgroup$ – Edward Nashton Sep 28 '15 at 5:26

Using closed form of $F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ where $\alpha=\frac{-1+\sqrt{5}}{2},\ \beta=\frac{-1-\sqrt{5}}{2}$ will work, but maybe after a long and tedious calculation. A simpler way is to look at it in the following way.

$$G_n=F_{n}F_{n+2}-F_{n+1}^2=F_n(F_n+F_{n+1})-F_{n+1}^2\\=F_n^2-F_{n+1}(F_{n+1}-F_n)=F_n^2-F_{n+1}F_{n-1}=-G_{n-1}\\\implies G_n=(-1)^{n-1}G_1=(-1)^{n-1}$$

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  • $\begingroup$ Thanks for the reply and the help. Would it be possible to get some clarification on how you got from $$(F_{n})^2-F_{n+1}*F_{n-1} to -G_{n-1}$$ and the lines after that. $\endgroup$ – Edward Nashton Sep 28 '15 at 5:51
  • $\begingroup$ I know how to get to the final line now and I understand why the solution is what it is, but I am unsure how you go from $G_n$ to $(-1)^{n-1}G_1$ algebraically from $G_n=-G_{n-1}$ $\endgroup$ – Edward Nashton Sep 28 '15 at 11:19
  • $\begingroup$ Note that $G_n=-G_{n-1}=(-1)^2G_{n-2}=(-1)^3G_{n-3}$ and so on. $\endgroup$ – Samrat Mukhopadhyay Sep 28 '15 at 16:16

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