power sums of fibonacci and lucas numbers

Quaestiones Mathematicae 34(2011), 75–83. c 2011 NISC Pty Ltd, www.nisc.co.za ⃝ DOI: 10.2989/16073606.2011.570298

POWER SUMS OF FIBONACCI AND LUCAS NUMBERS Wenchang Chu∗ School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, P.R. China. E-Mail [email protected]

Nadia N. Li School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, P.R. China. E-Mail [email protected]

Abstract. Polynomial representation formulae for power sums of the extended Fibonacci-Lucas numbers are established, which include, as special cases, four formulae for odd power sums of Melham type on Fibonacci and Lucas numbers, obtained recently by Ozeki and Prodinger (2009).

Mathematics Subject Classification (2000): Primary 11B39, Secondary 05A15. Key words: Fibonacci numbers, Lucas numbers, generating function, Stirling numbers of the first kind, Stirling numbers of the second kind.

1. Introduction and motivation. For two complex numbers a and c, define the following commonly extended Fibonacci-Lucas numbers {Gn }n≥0 by the recurrence relation Gn (a, c) = Gn−1 (a, c) + Gn−2 (a, c) (1) with the initial values being given by G0 (a, c) = a and G1 (a, c) = c.

(2)

They will be shortened as Gn = Gn (a, c). By means of the usual series manipulation (cf. [3] and [5, Section 1.13]), it is not hard to show the following facts. • Generating function g(x) :=

∞ ∑

Gk (a, c)xk =

k=0

=

a + (c − a)x 1 − x − x2

c − a + aα c − a + aβ − . (α − β)(1 − xα) (α − β)(1 − xβ)

(3a) (3b)

∗ Corresponding address: Dipartimento di Matematica, Universit´ a del Salento, Lecce-Arnesano, P.O. Box 193, Lecce 73100, Italy.

75

76

W. Chu and N.N. Li • Explicit formula Gn (a, c) =

uαn − vβ n α−β

with n ∈ N0 .

(4)

The well-known Fibonacci and Lucas numbers are the following particular cases Fn = Gn (0, 1) and Ln = Gn (2, 1). Throughout the paper, we utilize, for the sake of brevity, the following four symbols α :=

√ 1+ 5 2

and β :=

√ 1− 5 2

(5)

as well as u := c − a + aα

and v := c − a + aβ. (6) ∑n It is classically well-known that the arithmetic sum k=1 k m can be expressed as a polynomial of n. Suggested by Melham (cf. [4] and [9] also), Ozeki [7, Theorem 2] established recently the following interesting polynomial representation formula n ∑

1+2m F2k =

k=1

−

m ∑ i=0 m ∑ j=0

1+2i F1+2n

)( ) m ( ∑ 1 + 2m i + j + 1 (1 + 2j)(−5)i−m

m−j 2i + 1 ( ) (−1)m−j 1 + 2m F1+2j 5m m − j L1+2j j=i

(1 + i + j)L1+2j

(7a)

(7b)

where a misprint has been corrected. Motivated by this equation, the purpose of the present paper is to investigate the following polynomial representation for odd power sums of the extended Fibonacci-Lucas numbers: n ∑ k=1 n ∑ k=1

Gm 2k (a, c) =

m ∑

λi (a, c)Gi1+2n (a, c),

i=0

Gm 2k−1 (a, c) =

m ∑

µi (a, c)Gi2n (a, c).

i=0

The connecting coefficients λi (a, c) and µi (a, c) will be determined explicitly. After having submitted this paper, H. Prodinger informed us that there is another recent article [8] dealing with the same problem. Observe that there are two main differences between this paper and [8]. One is that we are working on the unified Fibonacci-Lucas numbers with two free parameters a and c. Consequently, our main theorems (cf. Theorems 4 and 5) contain four formulae due to Prodinger [8] as special cases. Another difference lies in the proving methods. Prodinger [8] employed principally the series rearrangements. Apart from that, the inverse series relations due Gould and Hsu [6] play the key role in our derivation.

Power sums of Fibonacci and Lucas numbers

77

2. Three preliminary relations. In order to investigate the power sums of the extended Fibonacci-Lucas numbers, we need first to show three preliminary results. Lemma 1. Let ε be an integer and m an odd natural number. Then the following summation formula holds Lm

n ∑

Gε+2km = Gε+m(2n+1) − Gε+m .

k=1

Proof.

According to the explicit formulae of Gn and Lm = αm + β m , we have uαε+m(2k+1) − vβ ε+m(2k+1) uαε+2km β m − vαm β ε+2km + α−β α−β ε+m(2k+1) ε+m(2k+1) ε+m(2k−1) uα − vβ uα − vβ ε+m(2k−1) = − α−β α−β

Lm Gε+2km =

where we have invoked the property αβ = −1. This leads us to the relation Lm Gε+2km = Gε+m(2k+1) − Gε+m(2k−1) . Then the formula stated in Lemma 1 follows immediately from the telescoping method. 2

Proposition 2. For m, n ∈ N0 , there holds the algebraic identity G1+2m n

−m

=5

m ∑

(−1)n(m−k)

( 1 + 2m )

k=0

×

k ∑ i=0

Proof.

(−1)i

(k) i

m−k

(a2 + ac − c2 )m−k

ai (a − 2c)i (a2 + c2 )k−i Gn−i+2nk .

Applying the binomial theorem, we get { G1+2m = n =

uαn vβ n − α−β α−β

1+2m ∑ k=0

}1+2m

(−1)k+1 ( 1 + 2m ) k nk 1+2m−k n(1+2m−k) u α v β . k (α − β)1+2m

Splitting the last sum into two parts and then making the replacement k → 1 + 2m − k for the summation index of the second sum, we can proceed with the

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W. Chu and N.N. Li

following computation {∑ m 2m+1 ∑ } 1+2m + Gn = k=0

k=m+1

(−1)k+1 ( 1 + 2m ) k nk 1+2m−k n(1+2m−k) u α v β k (α − β)1+2m

m { } ∑ (−1)nk+k ( 1 + 2m ) n 1+2m−2k k n 1+2m−2k = (uα ) (uv) − (vβ ) k (α − β)1+2m

=

k=0 m ∑ k=0

{ } (−1)(1+n)(m−k) ( 1 + 2m ) n 1+2k m−k n 1+2k (uα ) − (vβ (uv) ) . m−k (α − β)1+2m

According to (6), it is almost trivial to check that uv = c2 − a2 − ac and two further relations: k ( ) { }k ∑ k u2k = a2 + c2 + a(a − 2c)β = (a2 + c2 )k−i ai (a − 2c)i β i , i i=0

k ( ) { }k ∑ k v 2k = a2 + c2 + a(a − 2c)α = (a2 + c2 )k−i ai (a − 2c)i αi . i i=0

We can consequently express = G1+2m n

as follows:

m ∑ (−1)n(m−k) ( 1 + 2m )

(α − β)2m

k=0

×

Gn1+2m

k ∑

(−1)i

i=0

(k) i

m−k

(a2 + ac − c2 )m−k

ai (a − 2c)i (a2 + c2 )k−i

uαn−i+2nk − vβ n−i+2nk . α−β 2

This is equivalent to the formula stated in Proposition 2.

In 1973, Gould and Hsu [6] discovered the following fundamental pair of inverse series relations. Let {ak , bk }k≥0 be two sequences such that the ϕ-polynomials defined by ϕ(x; 0) ≡ 1

and ϕ(x; m) =

m−1 ∏

(ak + xbk )

with

m∈N

(8)

k=0

differ from zero for x, m ∈ N0 . Then there hold the inverse series relations f (m) = g(m) =

m ∑ k=0 m ∑ k=0

(−1)k (−1)k

(m)

ϕ(k; m)g(k),

(9a)

( m ) a + kb k k f (k). k ϕ(m; k + 1)

(9b)

k

For their applications to classical hypergeometric series evaluation and further development, refer to Chu [1, 2]. Now we are ready to show the following dual result of Proposition 2.


79

Proposition 3. For m, n ∈ N0 , there holds the algebraic identity m ∑

(−1)i

i=0

=

m ∑

(m) i

ai (a − 2c)i (a2 + c2 )m−i Gn−i+2nm

(−1)(m−k)(n+1)

(

k=0

Proof.

) m + k 1 + 2m k 2 5 (a + ac − c2 )m−k G1+2k . n 2k 1 + 2k

Define the shifted factorial by (x)0 = 1

and

(x)n = x(x + 1) · · · (x + n − 1) for n ∈ N.

By means of the binomial relation ( 1 + 2m ) m−k

=

k! (1 + 2m)! ( m ) 2 k (m!) (m + 1)k+1

we can reformulate the equation in Proposition 2 as ( ) 1 + 2k ∑ (−1)k+kn k! (−1)mn 5m (m!)2 G1+2m n k m = (−1) k (m + 1)k+1 (1+2k)(a2 +ac−c2 )k (1 + 2m)!(a2 + ac − c2 )m m

k=0

×

k ∑

(−1)i

(k) i

i=0

ai (a − 2c)i (a2 + c2 )k−i Gn−i+2nk .

The last equation matches (9b) exactly with the following specifications

f (k) = g(m) =

k (k) ∑ (−1)k+kn k! (−1)i ai (a − 2c)i (a2 + c2 )k−i Gn−i+2nk , 2 2 k i (1+2k)(a +ac−c ) i=0

(−1)mn 5m (m!)2 G1+2m n (1 + 2m)!(a2 + ac − c2 )m

and

ϕ(x; m) = (x + 1)m .

The dual relation corresponding to (9a) yields the following equation m ( ) ∑ (−1)m(n+1) m! i m (−1) ai (a − 2c)i (a2 + c2 )m−i Gn−i+2mn i (1 + 2m)(a2 + ac − c2 )m i=0

=

m ∑ k=0

(−1)k

(m) k

(k + 1)m

(−1)kn 5k (k!)2 G1+2k n . (1 + 2k)!(a2 + ac − c2 )k

This is, in turn, equivalent to the relation stated in Proposition 3.

2

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W. Chu and N.N. Li

3. Main theorems and corollaries. theorems of this paper.

Now we are ready to prove the two main

Theorem 4. (Polynomial representation for power sum of even order: m, n ∈ N0 ) n ∑

G1+2m 2j

j=1

m { m }∑ ∑ 1 + 2k ( 1 + 2m )( k + i ) (a2 +ac−c2 )m−i 1+2i 1+2i . = G1+2n − c 2i 1 + 2i m − k 5m−i L1+2k i=0 k=i

Proof. For the equation displayed in Proposition 2, replacing n by 2j and then summing across the resulting equation over 1 ≤ j ≤ n, we obtain n ∑

= 5−m G1+2m 2j

j=1

m ( ∑ 1 + 2m ) k=0

×

k ∑

(−1)i

i=0

m−k (k) i

(a2 + ac − c2 )m−k

ai (a − 2c)i (a2 + c2 )k−i

(10a) n ∑

G2j(1+2k)−i .

(10b)

j=1

The last sum with respect to j can be evaluated through Lemma 1 as { } L−1 1+2k G(1+2k)(2n+1)−i − G1+2k−i . Substituting this into (10b) and then applying Proposition 3 lead to another expression for (10b) L−1 1+2k

k ∑ 1 + 2k ( k + i ) i=0

2i

1 + 2i

{ } 1+2i 5i (a2 + ac − c2 )k−i G1+2i . 1+2n − G1

Combining this last expression with (10a) and interchanging the summation order result in the formula stated in Theorem 4. 2

Theorem 5. (Polynomial representation for power sum of odd order: m, n ∈ N0 ) n ∑

G1+2m 2j−1 =

j=1

m ∑

G1+2i 2n

i=0

−

m ∑ 1 + 2k ( 1 + 2m )( k + i ) (c2 −ac−a2 )m−i k=i

1 + 2i

m−k

2i

5m−i L1+2k

m a ∑( 1 + 2m ) (a2 + c2 )k 2 (c − a2 − ac)m−k . m−k 5m L1+2k k=0

Proof. Similarly for the equation displayed in Proposition 2, replacing n by 2j − 1 and then summing across the resulting equation over 1 ≤ j ≤ n, we get n ∑

−m G1+2m 2j−1 = 5

j=1

m ∑

(−1)m−k

k=0

×

k ∑ i=0

(−1)i

(

( 1 + 2m ) m−k

(a2 + ac − c2 )m−k

n ) ∑ k i a (a − 2c)i (a2 + c2 )k−i G(1+2k)(2j−1)−i . i j=1

(11a)

(11b)


81

The last sum with respect to j can analogously be evaluated by Lemma 1 as { } L−1 1+2k G2n(2k+1)−i − G−i . Substituting this into (11b) and then appealing to Proposition 3, we find the following alternative expression for (11b) ( k + i ) (1 + 2k)5i (a2 + c2 )k ∑ + (−1)k−i (a2 + ac − c2 )k−i G1+2i 2n . 2i L1+2k (1 + 2i)L 1+2k i=0 k

−a

Combining this expression with (11a) gives rise to the formula displayed in Theorem 5. 2 Finally as examples, we shall establish four polynomial representation formulae for Fibonacci and Lucas numbers by taking a = 0, c = 1 and a = 2, c = 1 in Theorem 4 and Theorem 5. Corollary 6. (Polynomial representation for powers of Fibonacci numbers) n ∑

1+2m F2j =

j=1

− n ∑

1+2m F2j−1 =

j=1

m ∑ i=0 m ∑ k=0 m ∑

1+2i F1+2n

m ∑ 1 + 2k ( 1 + 2m )( k + i ) (−5)i−m k=i

(−1)m−k 1+2i F2n

i=0

1 + 2i

m−k

( 1 + 2m ) F

1+2k

m−k

L1+2k

2i

L1+2k

5−m ;

m ∑ 1 + 2k ( 1 + 2m )( k + i ) 5i−m . 2i 1 + 2i m − k L1+2k k=i

We point out that the first identity is equivalent to (7a-7b) due to Ozeki [7]. Proof.

Specifying a = 0 and c = 1 in Theorem 4, we have n ∑ j=1

1+2m F2j =

m ∑

1+2i F1+2n

i=0

−(−5)−m

m ∑ 1 + 2k ( 1 + 2m )( k + i ) (−5)i−m k=i

1 + 2i

m−k

2i

L1+2k

m ( k ∑ 1 + 2m ) 1 + 2k ∑ (−5)i ( k + i ) . m − k L1+2k 2i 1 + 2i i=0

k=0

This leads to the first identity in view of the identity k ∑ (−5)i ( k + i ) F1+2k = (−1)k 2i 1 + 2i 1 + 2k i=0

(12)

which can be proved by means of the generating function method. The second identity follows directly by letting a = 0 and c = 1 in Theorem 5. 2

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W. Chu and N.N. Li

Corollary 7. (Polynomial representation for powers of Lucas numbers) n ∑ j=1 n ∑

L1+2m = 2j L1+2m 2j−1 =

j=1

m ∑ i=0 m ∑

L1+2i 1+2n L1+2i 2n

i=0 m ∑

−2

k=0

Proof.

m ∑ 1 + 2k ( 1 + 2m )( k + i ) k=i m ∑ k=i

1 + 2i

m−k

2i

m L−1 1+2k − 4 ;

1 + 2k ( 1 + 2m )( k + i ) (−1)i+m 2i 1 + 2i m − k L1+2k

(−1)m−k ( 1 + 2m ) . m−k L1+2k

Letting a = 2 and c = 1 in Theorem 4, we have n ∑

L1+2m = 2j

j=1

m ∑

L1+2i 1+2n

i=0

−

m ( ∑ k=0

m ∑ 1 + 2k ( 1 + 2m )( k + i ) k=i

1 + 2i

) 1 + 2m 1 + 2k m − k L1+2k

m−k k ∑ i=0

2i

L−1 1+2k

(k + i) 2i . 1 + 2i

Then the first identity in Corollary 7 follows because the last double sum results in 4m , which can be justified by utilizing first the identity k ∑ i=0

(k + i) 2i L1+2k = 1 + 2i 1 + 2k

(13)

and then the binomial sum m ( ) ∑ 1 + 2m k=0

m−k

= 4m .

The second identity in Corollary 7 can similarly be shown by specifying a = 2 and c = 1 in Theorem 5. 2 Before concluding the paper, it should be pointed out that the formulae displayed in Corollaries 6 and 7 coincide with four of eight identities collected in Sections 4 and 6 by Prodinger [8]. Acknowledgement. The authors are grateful to Professor H. Prodinger for having informed us of Reference [8]. References 1. W. Chu, Inversion techniques and combinatorial identities: a quick introduction to hypergeometric evaluations, Math. Appl. 283 (1994), 31–57.


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2. W. Chu, Inversion techniques and combinatorial identities: balanced hypergeometric series, Rocky Mountain J. Math. 32(2) (2002), 561–587. 3. W. Chu and V. Vincenti, Funzione generatrice e polinomi incompleti di Fibonacci e Lucas, Boll. Un. Mat. Ital. Serie VIII, B6(2) (2003), 289–308. 4. S. Clary and P.D. Hemenway, On sums of cubes numbers, Applications of Fibonacci Numbers 5 (1993), 123–136.

of

Fibonacci

5. L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and enlarged edition, D. Reidel Publishing Co., Dordrecht, 1974. 6. H.W. Gould and L.C. Hsu, Some new inverse series relations, Duke Math. J. 40 (1973), 885–891. 7. K. Ozeki, On Melham’s sum, Fibonacci Quart. 46/47(2) (2009), 107–110. 8. H. Prodinger, On a sum of Melham and its Variants, Fibonacci Quart. 46/47(3) (2009), 207–215. 9. M. Wiemann and C. Cooper, Divisibility of an F-L type convolution, Applications of Fibonacci Numbers 9 (2004), 267–287. Received 26 February, 2010 and in revised form 26 June, 2010.