Fibonacci Numbers and Their Lucas Coefficients

TL;DR

This paper proves a new identity linking Fibonacci and Lucas numbers, expressing Fibonacci numbers as a sum involving Lucas coefficients, with a detailed inductive proof and numeric illustration.

Contribution

It introduces a novel identity connecting Fibonacci and Lucas sequences, providing a detailed proof and practical example.

Findings

Established a new Fibonacci-Lucas identity for all n >= 2

Provided a rigorous inductive proof of the identity

Illustrated the identity with a numerical example

Abstract

We show that for the classical Fibonacci sequence (Fn) and the Lucas sequence (Ln) the following identity holds for every integer n >= 2: (n-1)Fn equals the sum from k=1 to n-1 of Lk multiplied by F(n-k). Equivalently, this gives a representation of the nth Fibonacci number as Fn = (1 / (n-1)) times the same sum. We present a detailed proof by mathematical induction and illustrate the identity with a numeric example.

Equations

$$(n-1)F_n=\sum _{k=1}^{n-1}{L}_k{F}_{n-k}.$$

$$\boxed{F_n=\frac{1}{n-1}\sum _{k=1}^{n-1}{L}_k{F}_{n-k}}$$

$$F_1=1,F_2=1,F_n=F_{n-1}+F_{n-2}(n\ge 3),$$

$$L_0=2,L_1=1,L_2=3,L_n=L_{n-1}+L_{n-2}(n\ge 2).$$

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

$$E^{(1)}:$$

$$E^{(2)}:$$

$$F_n=2F_{n-2}+F_{n-3},$$

$$E^{(3)}:$$

$$F_n=3F_{n-3}+2F_{n-4},$$

$$E^{(4)}:$$

$$F_n=5F_{n-4}+3F_{n-5},$$

$$⋮$$

$$\Sigma :=E^{(1)}+E^{(2)}+\cdots +E^{(n-1)},$$

$$\text{LHS}(\Sigma )=\underset{(n-1)\text{ times}}{\underbrace{F_n+F_n+\cdots +F_n}}=(n-1)F_n.$$

$$\Sigma \Rightarrow (n-1)F_n=\sum _{k=1}^{n-1}{a}_k{F}_{n-k}.$$

$$a_{k+1}=a_k+a_{k-1}(k\ge 2),$$

$$\sum _{k=1}^5{L}_k{F}_{6-k}=1\cdot F_5+3\cdot F_4+4\cdot F_3+7\cdot F_2+11\cdot F_1=5+9+8+7+11=40,$$

$$(n-1)F_n=\sum _{k=1}^{n-1}{L}_k{F}_{n-k}$$

$$\boxed{(n-1)F_n=\sum _{k=1}^{n-1}{L}_k{F}_{n-k}}.$$

$$\boxed{F_n=\frac{1}{n-1}\sum _{k=1}^{n-1}{L}_k{F}_{n-k}}.$$

$$S_n:=\sum _{k=1}^{n-1}{L}_k{F}_{n-k}.$$

$$S_2=\sum _{k=1}^1{L}_k{F}_{2-k}=L_1{F}_1=1\cdot 1=1,$$

$$S_m=(m-1)F_m\text{and}{S}_{m-1}=(m-2)F_{m-1}.$$

$$S_{m+1}=\sum _{k=1}^m{L}_k{F}_{m+1-k}.$$

$$S_{m+1}=L_1{F}_m+\sum _{k=2}^m{L}_k{F}_{m+1-k}=F_m+\sum _{j=1}^{m-1}{L}_{j+1}{F}_{m-j},$$

$$\sum _{j=1}^{m-1}{L}_{j+1}{F}_{m-j}=\sum _{j=1}^{m-1}(L_j+L_{j-1})F_{m-j}=\sum _{j=1}^{m-1}{L}_j{F}_{m-j}+\sum _{j=1}^{m-1}{L}_{j-1}{F}_{m-j}.$$

$$\sum _{j=1}^{m-1}{L}_{j-1}{F}_{m-j}=\sum _{i=0}^{m-2}{L}_i{F}_{m-1-i}=L_0{F}_{m-1}+\sum _{i=1}^{m-2}{L}_i{F}_{m-1-i}=L_0{F}_{m-1}+S_{m-1}.$$

$$S_{m+1}=F_{m}+S_{m}+\big{(}L_{0}F_{m-1}+S_{m-1}\big{)}.$$

$$S_{m+1}$$

$$=m{F}_m+m{F}_{m-1}$$

$$=m(F_m+F_{m-1})$$

$$=m{F}_{m+1}.$$

$$\boxed{F_n=\frac{1}{n-1}\sum _{k=1}^{n-1}{L}_k{F}_{n-k}}.$$

$$\boxed{\,F_{n}\;=\;\frac{1}{\,n-1\,}\sum_{k=1}^{\,n-1}\big{(}L_{k}\cdot F_{\,n-k}\big{)}\,}.$$

Full text

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Equivalently,

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We give a detailed proof by mathematical induction and present a small numeric example.

1 Preliminaries

We use the standard indexing

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for the Fibonacci numbers, and the Lucas numbers with the convenient extension

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(We include $L_{0}=2$ so that the Lucas recurrence is valid for all indices used below; note $L_{2}=L_{1}+L_{0}\Rightarrow 3=1+2$.)

2 The Work

We begin with the Fibonacci recurrence:

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2.1 Derivation by repeated substitution and summation

Here we show explicitly how repeated substitution of previous terms and then summing the expanded forms yields the identity in the statement.

Step 1 — single expansions. Start from (1) and perform repeated substitutions for $F_{n-1},F_{n-2},\dots$ to produce several different expanded forms of the same $F_{n}$. For clarity write the equation after $r-1$ substitutions as $E^{(r)}$ (so $E^{(1)}$ is the original relation).

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Each $E^{(r)}$ is simply the result of expanding $F_{n}$ by replacing the highest-index previous term repeatedly. After $r-1$ substitutions the right-hand side involves $F_{n-r+1},F_{n-r},\dots$ with integer coefficients.

Step 2 — collect several expanded copies. Now form the sum

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that is, add together the $n-1$ expanded forms of $F_{n}$ (from no substitutions up to $n-2$ substitutions). On the left-hand side each $E^{(r)}$ contributes a copy of $F_{n}$, so

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On the right-hand side we collect the contributions to each $F_{n-k}$ (for $k=1,\dots,n-1$) coming from all expansions $E^{(r)}$. Denote by $a_{k}$ the total coefficient of $F_{n-k}$ appearing in the sum $\Sigma$. By construction,

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Step 3 — the recurrence for the collected coefficients. Inspecting how coefficients accumulate when performing one further substitution shows that the sequence $(a_{k})_{k\geq 1}$ satisfies the same two-term recurrence as the Lucas numbers. Concretely, each time we perform one more substitution (moving from $E^{(r)}$ to $E^{(r+1)}$) the vector of coefficients shifts and new coefficients are obtained by summing adjacent previous coefficients. Therefore the collection $a_{k}$ obeys

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with initial values $a_{1}=1$ (from each expansion there is exactly one $F_{n-1}$ appearing in total) and $a_{2}=3$ (which can be checked from the first two or three expansions). These initial values match the Lucas sequence seed $L_{1}=1,L_{2}=3$. Thus $a_{k}=L_{k}$ for all $k\geq 1$.

Illustration (numerical check). Take $n=6$. The expansions $E^{(1)},\dots,E^{(5)}$ produce collected coefficients $a_{1},\dots,a_{5}=1,3,4,7,11$, which are the Lucas numbers $L_{1},\dots,L_{5}$. Summing those coefficients times the corresponding Fibonacci numbers gives

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and the left-hand side is $(6-1)F_{6}=5\cdot 8=40$, as required.

This constructive procedure (expand, collect, sum the expanded forms) explains how the identity

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was discovered: repeated substitution produces multiple expanded expressions of $F_{n}$; summing those $n-1$ expressions yields $(n-1)F_{n}$ on the left and produces Lucas-number weights on the earlier Fibonacci terms on the right.

3 Main theorem and detailed proof by induction

Theorem 1**.**

For every integer $n\geq 2$,

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Equivalently,

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Proof.

Define

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We will prove $S_{n}=(n-1)F_{n}$ for all $n\geq 2$ by induction on $n$.

Base case: $n=2$. Then

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and $(2-1)F_{2}=1\cdot 1=1$. So the claim holds for $n=2$.

Inductive hypothesis: Suppose for some integer $m\geq 2$ we have

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We prove the statement for $n=m+1$.

Consider

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Separate the first term (when $k=1$) and reindex the remaining sum:

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where the change of variable $j=k-1$ was used in the second sum.

Use the Lucas recurrence $L_{j+1}=L_{j}+L_{j-1}$ (valid for $j\geq 1$, and recall $L_{0}=2$). Thus

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The first sum on the right is exactly $S_{m}$. For the second sum, reindex $i=j-1$, which yields

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Combining these pieces gives

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Now substitute the inductive hypotheses $S_{m}=(m-1)F_{m}$ and $S_{m-1}=(m-2)F_{m-1}$, and recall $L_{0}=2$. We obtain

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This proves $S_{m+1}=mF_{m+1}$, completing the inductive step.

By induction, the identity $S_{n}=(n-1)F_{n}$ holds for all integers $n\geq 2$. Rearranging yields the equivalent form

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∎

4 Result (bold statement)

For every integer $n\geq 2$, the $n$-th Fibonacci number can be written as

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5 Conclusion

We proved that repeated expansion and collection of terms leads to the identity $(n-1)F_{n}=\sum_{k=1}^{n-1}L_{k}F_{n-k}$. Dividing both sides by $n-1$ gives a representation of $F_{n}$ as a weighted sum of earlier Fibonacci numbers where the weights are Lucas numbers and the overall factor is $1/(n-1)$. This connects the two classical sequences in a simple algebraic identity. Possible extensions include studying analogous identities for generalized Fibonacci sequences (different seeds) or for higher-order recurrences (Tribonacci, etc.).