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Generalized Fibonacci Sequence

What is the formula for the generalized Fibonnaci sequence,

\(F_{n + k} = F_{n + k - 1} + F_{n + k - 2} + ... + F_{n}\), where \(k \in \mathbb{N}\)

Note by Zi Song Yeoh
4 years, 7 months ago

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You can solve using the characteristic equation (just wiki recurrence relations). Note that the sequence's general formula is dependent on what you choose for F1, F2 ... Fk - based on what the Fi are.

Gabriel Wong - 4 years, 7 months ago

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Indeed. Learn how to solve Linear Recurrence Relations using the characteristic equations.

The characteristic equation is \( x^k = x^{k-1} + x^{k-2} + \ldots + x^0\), which has \(k\) roots of the form \(\{ \alpha_i\}_{i=1}^k \). Then, your sequence will have the value \( F_n = \sum A_i (\alpha_i)^n \), where \(A_i\) depends on the starting values that you chose. [Note: I glossed over the case of repeated roots, which have to be treated differently].

For example, when \( k = 2 \), then the equation is \( x^2 = x + 1 \) or equivalently \( x^2 - x - 1=0\). This has roots \( \frac {1 \pm \sqrt{5} } {2} \), and appear as the powers in Binet's formula.

Calvin Lin Staff - 4 years, 7 months ago

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The formula involves the square root of 5, which I write as sqrt[5]. The nth Fibonacci is:

 a*x^n + (1-a)*y^n

where a = (3+sqrt[5])/(5+sqrt[5]) x = (1+ sqrt[5])/2 y = (1- sqrt[5])/2

So my challenges to you are: first, to prove that the formula works, and second, to derive the equation you want for the sum of the first n terms, using the formula for the sum of a geometric series.

Superman Son - 4 years, 7 months ago

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You can use Binet's Formula to find the n-th Fibonacci number. It is easy to prove by induction.

Zi Song Yeoh - 4 years, 7 months ago

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I know the formula for the nth Fibonacci number, I'm asking for the formula for the recurrence stated. (including Fibonacci Sequence)

Zi Song Yeoh - 4 years, 7 months ago

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