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International Mathematical Forum, 5, 2010, no. 3, 99 - 107

Fibonacci and Lucas Sums by Matrix Methods

Bahar Demirturk

Sakarya University Faculty of Science and ArtsDepartment of Mathematics, Sakarya, Turkey 54187

[email protected]

Abstract

In this paper, some Fibonacci and Lucas sums are derived by using

the matrices S =

1/2 5/21/2 1/2

and K =

0 5

1 0

. The most notable

side of this paper is our proof method, since all the identities used in

the proofs of main theorems are proved previously by using the matricesS and K. Although the identities we proved are known, our proofs arenot encountered in the Fibonacci and Lucas numbers literature.

Mathematics Subject Classification: 11B37, 11B39, 40C05

Keywords: Fibonacci numbers; Lucas numbers; Fibonacci matrix

1 Introduction

The Fibonacci sequence {Fn} is defined by the recurrence relation Fn = Fn1+

Fn2, for n 2 with F0 = 0 and F1 = 1. The Lucas sequence {Ln} , consideredas a companion to Fibonacci sequence, is defined recursively by Ln = Ln1 +Ln2, for n 2 with L0 = 2 and L1 = 1. It is well known that Fn = (1)

n+1Fnand L

n = (1)nLn, for every n Z. For more detailed information see

[9],[10].This paper presents an interesting investigation about some special rela-

tions between matrices and Fibonacci, Lucas numbers. This investigation isvaluable, since it provides students to use their theoretical knowledge to ob-tain new Fibonacci and Lucas identities by different methods. So, this papercontributes to Fibonacci and Lucas numbers literature, and encourage many

researchers to investigate the properties of such number sequences.

2 Main Theorems

Lemma 1 If X is a square matrix with X2 = X+ I, then Xn = FnX+ Fn1Ifor all n Z.


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100 B. Demirturk

Proof. If n = 0, then the proof is obvious. It can be shown by induction thatXn = FnX+Fn1I for every n N. We now show that X

n = FnX+Fn1I

for every n N. Let Y = I X = X1. Then

Y2 = (IX)2 = I 2X+ X2 = I 2X+ X+ I = I+ I X = I+ Y.

This shows that Yn = FnY + Fn1I. That is, (X1)

n= Fn (I X) + Fn1I.

Therefore (1)n Xn = FnX+ (Fn + Fn1) I = FnX+ Fn+1I. Thus

Xn = (1)n+1 FnX+ (1)n Fn+1I = FnX+ Fn1I

We can give Corollary1 from Lemma1 easily. Also one can consult [9],[10] formore information about the matrice Q.

Corollary 1 LetQ = 1 1

1 0

. Then Qn

= Fn+1 Fn

Fn Fn1

for every n Z.

Corollary 2 Let S =

1/2 5/21/2 1/2

. Then Sn =

Ln/2 5Fn/2Fn/2 Ln/2

for every

n Z.

Let us give the following corollary for using in the next theorems.

Lemma 2 Let A =

a bc d

and K = S + S1 =

0 51 0

. Then KA =

5c 5da b

.

Lemma 3 L2n 5F2n = 4(1)

n for all n Z.

Proof. Since det S = 1, det(Sn) = (det S)n = (1)n. Moreover, since

Sn =

Ln/2 5Fn/2Fn/2 Ln/2

, we get det Sn = (L2n 5F

2n)/4. Thus it follows that

L2n 5F2

n = 4(1)n (2.1)

Let us give a different proof of one of the fundamental identities of Fibonacciand Lucas numbers, by using the matrix S.

Lemma 4 2Ln+m = LnLm +5FnFm and 2Fn+m= FnLm+LnFm for alln, m

Z.


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Fibonacci and Lucas sums by matrix methods 101

Proof. SinceLn/2 5Fn/2Fn/2 Ln/2

Lm/2 5Fm/2Fm/2 Lm/2

= SnSm = Sn+m =

Ln+m/2 5Fn+m/2Fn+m/2 Ln+m/2

,

it is seen that(LnLm + 5FnFm)/4 5(FnLm + LnFm)/4(FnLm + LnFm)/4 (LnLm + 5FnFm)/4

=

Ln+m/2 5Fn+m/2Fn+m/2 Ln+m/2

.

Thus it follows that

2Ln+m = LnLm + 5FnFm (2.2)

2Fn+m = FnLm + LnFm (2.3)

Lemma 5 2(1)mLnm = LmLn5FmFn and 2 (1)m Fnm = FnLmLnFm

for all n, m Z.

Proof. Since

Snm = SnSm = Sn(Sm)1 = Sn(1)m

Lm/2 5Fm/2Fm/2 Lm/2

= (1)m

Ln/2 5Fn/2Fn/2 Ln/2

Lm/2 5Fm/2Fm/2 Lm/2

= (1)m

(LmLn 5FmFn)/4 5(LnFm FnLm)/4(LnFm FnLm)/4 (LmLn 5FmFn)/4

and

Snm =

Lnm/2 5Fnm/2Fnm/2 Lnm/2

,

then we get

2(1)mLnm = LmLn 5FmFn

and

2 (1)m Fnm = FnLm LnFm

Lemma 6 LmLn = Ln+m + (1)mLnm and LmFn = Fn+m + (1)

mFnm forall n, m Z.


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102 B. Demirturk

Proof. By the definition of the matrix Sn, it can be seen that

Sn+m + (1)mSnm =

(Ln+m + (1)

mLnm) /2 5 (Fn+m + (1)mFnm) /2

(Fn+m + (1)mFnm) /2 (Ln+m + (1)

mLnm) /2

.

On the other hand,

Sn+m + (1)mSnm = SnSm + (1)mSnSm = Sn

Sm + (1)m Sm

=

Ln/2 5Fn/2Fn/2 Ln/2

Lm/2 5Fm/2Fm/2 Lm/2

+ (1)m

Lm/2 5Fm/2

Fm/2 Lm/2

=

Ln/2 5Fn/2Fn/2 Ln/2

Lm 0

0 Lm

=

LmLn/2 5LmFn/2LmFn/2 LmLn/2

.

Then the result follows.

Theorem 2.1 Let n N and m, k Z with m = 0. Then

nj=0

Lmj+k =Lk Lmn+m+k + (1)

m(Lmn+k Lkm)

1 + (1)m Lm

and

n

j=0

Fmj+k =Fk Fmn+m+k + (1)

m(Fmn+k Fkm)

1 + (1)m Lm.

Proof. It is known that

I (Sm)n+1 = (I Sm)n

j=0

(Sm)j .

If det(I Sm) = 0, then we can write

(I Sm)1

I (Sm)n+1

Sk =

n

j=0

Smj+k =

1

2

nj=0

Lmj+k5

2

nj=0

Fmj+k

1

2

nj=0Fmj+k

1

2

nj=0Lmj+k

(2.4)

From Lemma3, it is easy to see that

det(I Sm) = (1 Lm/2)2 5F2m/4 = 1 + (1)

m Lm = 0


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for m = 0. If we take D = 1 + (1)m Lm, then we get

(I Sm)1 =1

D

1 Lm/2 5Fm/2

Fm/2 1 Lm/2

=

1

D

1

Lm2

I+

Fm2 K

.

Thus it is seen that

(I Sm)1(Sk Smn+m+k) =1

D

1

Lm2

I +

Fm2

K

(Sk Smn+m+k)

=1

D

1

Lm2

(Sk Smn+m+k) +

Fm2

K(Sk Smn+m+k)

(2.5)

By Corollary1 and Lemma2, we get

K(Sk Smn+m+k) =

5(Fk Fmn+m+k)/2 5(Lk Lmn+m+k)/2(Lk Lmn+m+k)/2 5(Fk Fmn+m+k)/2

(2.6)

Using (2.6) in (2.5), we obtain

(I Sm)1(Sk Smn+m+k)

=1

D

1

Lm2

(Lk Lmn+m+k) /2 5 (Fk Fmn+m+k) /2(Fk Fmn+m+k) /2 (Lk Lmn+m+k) /2

+Fm

2

5(Fk Fmn+m+k)/2 5(Lk Lmn

+m+k)/2(Lk Lmn+m+k)/2 5(Fk Fmn+m+k)/2

.

Thus, from the identity (2.4), it follows that

nj=0

Lmj+k =1

D

1

Lm2

(Lk Lmn+m+k) +

5Fm2

(Fk Fmn+m+k)

=1

D

Lk Lmn+m+k

LmLk 5FmFk2

+LmLmn+m+k 5FmFmn+m+k

2

(2.7)

By Lemma5, (2.7) becomes

nj=0

Lmj+k =Lk Lmn+m+k + (1)

m(Lmn+k Lkm)

1 + (1)m Lm.


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104 B. Demirturk

On the other hand, by (2.5) and (2.6) we get

nj=0

Fmj+k =1

D

1

Lm2

(Fk Fmn+m+k) +

Fm2

(Lk Lmn+m+k)

=1

D

Fk Fmn+m+k

FkLm LkFm2

+Fmn+m+kLm Lmn+m+kFm

2

(2.8)

By Lemma5, it follows that

nj=0

Fmj+k =Fk Fmn+m+k + (1)

m(Fmn+k Fkm)

1 + (1)m Lm

Theorem 2.2 Let n N and m, k Z. Then

n

j=0

(1)jLmj+k =Lk + Lmn+m+k + (1)

m(Lmn+k + Lkm)

1 + (1)m

+ Lm

and

nj=0

(1)jFmj+k =Fk + Fmn+m+k + (1)

m(Fmn+k + Fkm)

1 + (1)m + Lm.

Proof. We prove the theorem in two phases, by taking n as an even and oddnatural number. Firstly assume that n is an even natural number. Then

I+ (Sm)n+1 = (I + Sm)n

j=0(1)j (Sm)j .

Since det(I+ Sm) = 1 + (1)m + Lm = 0, by Lemma3, we can write

(I+ Sm)1

I + (Sm)n+1

Sk =n

j=0

(1)jSmj+k

=

1

2

nj=0

(1)jLmj+k5

2

nj=0

(1)jFmj+k

1

2

nj=0

(1)jFmj+k1

2

nj=0

(1)jLmj+k

(2.9)

Taking d = 1 + (1)m + Lm, we get

(I + Sm)1 =1

d

1 + Lm/2 5Fm/2Fm/2 1 + Lm/2

=1

d

1 +

Lm2

I

Fm2

K

.


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Thus it is seen that

(I+ Sm)1

Sk + Smn+m+k

=1

d

(1 +

Lm2

)IFm2

K

Sk + Smn+m+k

(2.10)

By Corollary2 and Lemma2, it is seen that

K

Sk + Smn+k

=

5(Fk + Fmn+m+k)/2 5(Lk + Lmn+m+k)/2(Lk + Lmn+m+k)/2 5(Fk + Fmn+m+k)/2

(2.11)

Using (2.11) in (2.10), we get

(I + Sm)1

Sk + Smn+m+k

=1

d

1 +

Lm2

(Lk + Lmn+m+k) /2 5 (Fk + Fmn+m+k) /2(Fk + Fmn+m+k) /2 (Lk + Lmn+m+k) /2

Fm

2 5(Fk + Fmn+m+k)/2 5(Lk + Lmn+m+k)/2

(Lk + Lmn+m+k)/2 5(Fk + Fmn+m+k)/2

.

Then, by the identity (2.9), it follows that

nj=0

(1)jLmj+k =1

d

(1 +

Lm2

)(Lk + Lmn+m+k) 5Fm

2(Fk + Fmn+m+k)

=1

d

Lk + Lmn+m+k +

(LmLk 5FmFk) + (LmLmn+m+k 5FmFmn+m+k)

2

.

By Lemma5 and the fact that d = 1 + (1)m + Lm, we obtain

nj=0

(1)jLmj+k =Lk + Lmn+m+k + (1)

m(Lmn+k + Lkm)

1 + (1)m + Lm(2.12)

Similarly it can be easily seen that

nj=0

(1)jFmj+k =Fk + Fmn+m+k + (1)

m(Fmn+k + Fkm)

1 + (1)m + Lm(2.13)

by Lemma5.Now assume that n is an odd natural number. Hence we get

nj=0

(1)jLmj+k =n1j=0

(1)jLmj+k Lmn+k (2.14)


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106 B. Demirturk

Since n is an odd natural number then n 1 is even. Thus taking n 1 in(2.12) and using it in (2.14), it follows that

n

j=0(1)jLmj+k =

Lk + Lmn+k + (1)m(Lmn+km + Lkm)

1 + (1)m + Lm Lmn+k

=Lk + (1)

m(Lmn+km + Lkm) (1)mLmn+k LmLmn+k

1 + (1)m + Lm(2.15)

Using Lemma6 in (2.15), we get

nj=0

(1)jLmj+k =Lk Lmn+m+k + (1)

m(Lkm Lmn+k)

1 + (1)m + Lm.

In a similar way, it can be seen that

nj=0

(1)jFmj+k =n1j=0

(1)jFmj+k Fmn+k (2.16)

Then using (2.13) in (2.16), it follows that

nj=0

(1)jFmj+k =Fk + Fmn+k + (1)

m(Fmn+km + Fkm)

1 + (1)m + Lm Fmn+k

=Fk + (1)

m(Fmn+km + Fkm) (1)mFmn+k LmFmn+k

1 + (1)m + Lm.

Thus by Lemma6, we obtain the expected formula

nj=0

(1)jFmj+k =Fk Fmn+m+k + (1)

m(Fkm Fmn+k)

1 + (1)m + Lm

The author would like to thank Professor Refik Keskin for the contributionsto this paper.

References[1] D. Kalman and R. Mena, The Fibonacci Numbers-Exposed, Mathematics

Magazine, Volume 76 (2003), 167 - 181.

[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of NumbersOxfordUniversity Press, USA, 1980.


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[3] J. R. Silvester, The r-subsequences of the Fibonacci sequences, MathsGazette, 90 (2006), 263-266.

[4] J. R. Silvester, Fibonacci properties by matrix methods, MathematicalGazette, 63 (1979), 188-19.

[5] K.W. Yang, Fibonacci with golden ring, Mathematics Magazine 70 (1997),131-135.

[6] M. Z. Spiyev, Fibonacci Identities via the Determinant sum Property,College Mathematics Journal, 37:4 (2006), 286-289.

[7] R. C. Johnson, Matrix methods for Fibonacci and related sequences,www.dur.ac.uk/bob.johnson/fibonacci/.

[8] R. Keskin and B. Demirturk, Some New Fibonacci and Lucas Identitiesby Matrix Methods, International Journal of Mathematical Education In

Science and Technology, accepted for publication.

[9] S. Vajda, Fibonacci and Lucas numbers and the golden section, Ellis Hor-wood Limited Publ., England, 1989.

[10] T. Koshy, Fibonacci and Lucas numbers with applications, John Wileyand Sons, Proc., New York-Toronto, 2001.

Received: June, 2009

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