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While recently revisiting an old Project Euler problem, which states:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

I noticed the following.

Let $F_n = 1, 2, 3, 5, 8, \ldots$ as defined in the problem. Then, assuming zero indexing, for all $n \ge 2$,

$$F_n^3 + F_{n-1}^3 - F_{n-2}^3 = F_{3n+1}.$$

This evidently holds for the first few $n$:

$$F_2^3 + F_1^3 - F_0^3 = 27 + 8 - 1 = 34 = F_7$$ $$F_3^3 + F_2^3 - F_1^3 = 125 + 27 - 8 = 144 = F_{10}$$

and so forth.

I have tried using Cassini's Identity and its generalizations to prove this, but to no avail.

I've also found this answer, but cannot find the derivation of this solution, and cannot quite reconcile it with my observation.

My question: does my observation hold true, and if so, how can I go about proving this. What am I missing?

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  • $\begingroup$ For the original problem, you maya also note that the even valued fibos obey $x_{n+1}=4x_n+x_{n-1}$ $\endgroup$ – Hagen von Eitzen Feb 7 at 22:15
  • $\begingroup$ shouldn't it be $F_n^3 + F_{n-1}^3 - F_{n-2}^3 = F_{3(n-1)}$ which is valid for whichever $n$, even negative ? $\endgroup$ – G Cab Feb 8 at 0:08
  • $\begingroup$ yes, i made a typo $\endgroup$ – Emily Mar 4 at 20:22
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We have Binet's formula, $$ F_n=\frac{\phi^n-\psi^n}{\sqrt 5}$$ where $\phi=\frac{1+\sqrt 5}2$ and $\psi=\frac{1-\sqrt 5}2$ are the two roots of $X^2-X-1$. You can plug this into your conjectured equality, expand and simplify (using in particular that $\phi+\psi=1$, $\phi-\psi=\sqrt 5$, $\phi\psi=-1$), and check.

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  • $\begingroup$ of course, i should have seen this $\endgroup$ – Emily Feb 7 at 22:36

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