RELATIONS BETWEEN THE RECIPROCAL SUM AND THE ALTERNATING SUM FOR
GENERALIZED LUCAS NUMBERS
XIAOLI YE and ZHIZHENG ZHANG
Abstract. We establish relations between reciprocal and alternating sums involving generalized Lucas numbers.
1.
Introduction
Let p be a nonzero real number. The generalized Fibonacci and Lucas numbers are defined by
(1)
U0 = 0,
U1 = 1,
Un+2 = pUn+1 + Un ,
(n ≥ 0)
(2)
V0 = 2,
V1 = p,
Vn+2 = pVn+1 + Vn ,
(n ≥ 0)
respectively. Let α and β be the roots of x2 − px − 1 = 0. Then we have the Binet’s formulas:
αn − β n
(3)
Un =
,
α−β
(4)
Vn = α n + β n .
For p = 1, {Un } and {Vn } are the well-known Fibonacci numbers Fn and Lucas numbers Ln , respectively.
Received March 3, 2006.
2000 Mathematics Subject Classification. Primary 11B39, 40A99.
Key words and phrases. Generalized Lucas number, reciprocal sum, alternating sum.
This research is supported by the National Natural Science Foundation of China (Grant No. 10471016), the Natural Science
Foundation of Henan Province (Grant No. 0511010300) and the Natural Science Foundation of the Education Department of Henan
Province (Grant No. 200510482001).
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In [2], S. Rabinowitz discusses algorithmic aspects of certain finite reciprocal sums and propose several open
problems involving Fibonacci and Lucas numbers. Melham [1] gave several relations between the reciprocal sum
and the alternating sum for generalized Fibonacci numbers, which we subsequently extended in [4]. In this note
we establish some relations between the reciprocal sum and the alternating sum for generalized Lucas numbers.
For Un and Vn , we have the following well known expansions:
(5)
Un+1 =
[ n2 ] X
n−k
k=0
k
pn−2k ,
(n ≥ 0),
and
(6)
Vn =
[ n2 ]
X
k=0
n − k n−2k
n
p
,
n−k
k
(n ≥ 1).
See [3, (2.7),(2.8)]. From (5) and (6), since p is a nonzero real number, it is easy to obtain that Un > 0 (n ≥ 1)
and Vn > 0 (n ≥ 0) for p > 0. For p < 0, if n is an odd number, then we have Un+1 < 0 (n ≥ 0) and Vn < 0
(n ≥ 1). If n is an even number, then we have Un+1 > 0 (n ≥ 0) and Vn > 0 (n ≥ 0). Hence we have Un 6= 0
(n ≥ 1) and Vn 6= 0 (n ≥ 0). Our main result is the following.
Theorem 1.1. Let m and k be positive integers. Put
(7)
Sk (m) =
∞
X
1
n=1
Vn Vn+k Vn+2k . . . Vn+mk
and
(8)
Tk (m) =
∞
X
(−1)n−1
.
V V
V
. . . Vn+mk
n=1 n n+k n+2k
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Then
Sk (m) =
(9)
"
1
1 + (−1)(m−1)k − V(m+1)k
·
k
X
(−1)(m−1)k Vi+(m+1)k − Vi+2(m+1)k
i=1
Vi Vi+k Vi+2k . . . Vi+(m+1)k
+ (p2 + 4)U(m+1)k U(m+2)k Tk (m + 2)
i
and
(10)
( k
X
1
(−1)i−1
Tk (m) =
1 + (−1)(m−1)k − (−1)k V(m+1)k i=1 Vi Vi+k Vi+2k . . . Vi+(m+1)k
h
i
· (−1)(m−1)k Vi+(m+1)k − (−1)k Vi+2(m+1)k
o
+(p2 + 4)U(m+1)k U(m+2)k Sk (m + 2) .
When p = 1, by special choices of k and m, we have the following interesting results:
∞
X
1
L
L
n
n+1
n=1
=
∞
X
7
(−1)n−1
− 10
;
12
L L
L
L
n=1 n n+1 n+2 n+3
∞
X
(−1)n−1
L L
n=1 n n+1
=
∞
X
1
1
+2
;
4
L
L
L
n
n+1
n+2 Ln+3
n=1
=
∞
3
15 X
(−1)n−1
−
;
28
2 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4
∞
X
1
n=1
Ln Ln+1 Ln+2
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∞
X
=
∞
1
11
15 X
+
;
168
2 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4
=
∞
X
(−1)n−1
13
− 15
;
924
L L
L
L
L
L
n=1 n n+1 n+2 n+3 n+4 n+5
=
∞
29
25 X
1
+
;
2772
3 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4 Ln+5
1
L
L
n=1 n n+2
=
∞
X
139
(−1)n−1
− 24
;
396
L L
L
L
n=1 n n+2 n+4 n+6
∞
X
(−1)n−1
L L
n=1 n n+2
=
(−1)n−1
L L
L
n=1 n n+1 n+2
∞
X
1
L
L
L
L
n=1 n n+1 n+2 n+3
∞
X
(−1)n−1
L L
L
L
n=1 n n+1 n+2 n+3
∞
X
∞
X
1
L
L
L
n=1 n n+2 n+4
∞
X
95
1
− 242
;
396
L
L
L
L
n=1 n n+2 n+4 n+6
∞
123
199
105 X
(−1)n−1
=
+
−
;
16 · 11 · 29 84 · 18 · 47
2 n=1 Ln Ln+2 Ln+4 Ln+6 Ln+8
∞
X
∞
(−1)n−1
123
199
105 X
1
=
−
−
.
L L
L
16 · 11 · 29 84 · 18 · 47
2 n=1 Ln Ln+2 Ln+4 Ln+6 Ln+8
n=1 n n+2 n+4
2.
The proof of the results
Lemma 2.1. We have
(11)
Vn+k + (−1)k Vn−k = Vn Vk ;
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(12)
Vn+k = Vn Uk+1 + Vn−1 Uk ;
(13)
Vn+k−1 =
(14)
U2k = Uk Vk ;
(15)
Un+k = Un Uk+1 + Un−1 Uk ;
(16)
2
− Umk U(m+2)k = (−1)mk Uk2 ;
U(m+1)k
(17)
2
− Vn+mk Vn+(m+2)k = −(−1)n+mk (p2 + 4)Uk2 ;
Vn+(m+1)k
(18)
Uk Vn−mk = (−1)mk (Vn Umk+k − Vn+k Umk );
(19)
U(m+2)k Vi+(m+1)k − (−1)k U(m+1)k Vi+mk = Uk Vi+2(m+1)k ;
(20)
Un+k − (−1)k Un−k = Vn Uk ;
Vn Vn+(m+1)k = −
(21)
Uk−1 Vn+k + (−1)k−1 Vn
;
Uk
(−1)(m+1)k
(−1)(m+1)k
U(m+1)k Vn+(m+1)k Vn+(m+2)k +
U(m+2)k Vn+mk Vn+(m+2)k
Uk
Uk
− (−1)n+k (p2 + 4)Uk U(m+2)k ;
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Vn+mk Vn+(m+2)k − Vn Vn+(m+1)k =
(−1)(m+1)k
U(m+1)k Vn+(m+1)k Vn+(m+2)k
Uk
(−1)(m+1)k
+ 1−
U(m+2)k Vn+mk Vn+(m+2)k
Uk
(22)
+ (−1)n+k (p2 + 4)Uk U(m+2)k ;
Vn+mk Vn+(m+2)k + (−1)k−1 Vn Vn+(m+1)k =
(23)
(−1)mk
U(m+1)k Vn+(m+1)k Vn+(m+2)k
Uk
(−1)mk
U(m+2)k Vn+mk Vn+(m+2)k
+ 1−
Uk
+ (−1)n (p2 + 4)Uk U(m+2)k .
Proof. Use the Binet formulas (3) and (4) of Un and Vn .
Lemma 2.2. We have
(24)
∞
X
1
n=1
Vn Vn+k . . . Vn+(m−1)k Vn+(m+1)k
=
k
Umk + (−1)mk Uk
(−1)mk Uk X
1
Sk (m) −
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
and
∞
X
(25)
(−1)n−1
Umk + (−1)(m+1)k Uk
=
Tk (m)
V V
. . . Vn+(m−1)k Vn+(m+1)k
U(m+1)k
n=1 n n+k
−
k
(−1)i−1
(−1)(m+1)k Uk X
.
U(m+1)k
VV
. . . Vi+mk
i=1 i i+k
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Proof. We give only the proof of (24), the proof of (25) is similar. Using (11), we have
∞
X
1
V
V
.
.
.
V
n+(m−1)k Vn+(m+1)k
n=1 n n+k
=
∞
X
Vn+(m+1)k + (−1)k Vn+(m−1)k
Vn Vn+k . . . Vn+(m−1)k Vn+(m+1)k
n=1
=
∞ X
n=1
=
!
1
Vk Vn+mk
(−1)k
1
+
Vn Vn+k . . . Vn+(m−1)k
Vn Vn+k . . . Vn+(m−2)k Vn+(m+1)k
1
Vk Vn+mk
∞
(−1)k X
1
1
Sk (m) +
.
Vk
Vk n=1 Vn Vn+k . . . Vn+(m−2)k Vn+mk Vn+(m+1)k
Using (12), (13) and (15), we have
∞
X
1
V
V
.
.
.
V
n+(m−2)k Vn+mk Vn+(m+1)k
n=1 n n+k
=
=
∞
X
Vn+(m−2)k
Vn+(m+1)k + (−1)k UU2k
k
n=1
Vn Vn+k . . . Vn+(m−2)k Vn+mk Vn+(m+1)k
∞ X
n=1
!
Uk
U3k Vn+(m−1)k
1
Vn Vn+k . . . Vn+(m−2)k Vn+mk
+
!
(−1)k UU2k
k
Vn Vn+k . . . Vn+(m−3)k Vn+mk Vn+(m+1)k
·
Uk
U3k Vn+(m−1)k
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∞
Uk
1
(−1)k U2k X
=
Sk (m) +
.
U3k
U3k
V
V
.
.
.
V
V
n+(m−3)k n+(m−1)k Vn+mk Vn+(m+1)k
n=1 n n+k
Similarly, we obtain
∞
X
1
V V
. . . Vn+(m+1)k
n=1 n n+2k
∞
(−1)k Umk X
1
=
Sk (m) +
U(m+1)k
U(m+1)k n=1 Vn+k . . . Vn+(m+1)k
Uk
=
Uk
U(m+1)k
Sk (m) +
k
(−1)k Umk
(−1)k Umk X
1
Sk (m) −
.
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
Hence, using (16) and (17), we have
∞
X
1
V
V
.
.
.
V
n+(m−1)k Vn+(m+1)k
n=1 n n+k
(−1)k Uk
(−1)2k Uk U2k
(−1)mk Uk U2k
1
=
1+
+
+ ... +
Vk
U3k
U3k U4k
Umk U(m+1)k
k
(−1)mk U2k
(−1)mk U2k X
1
+
Sk (m) −
U(m+1)k
Vk U(m+1)k i=1 Vi Vi+k . . . Vi+mk
U(m−1)k
1
Uk U2k
Uk
U2k
U2k
U3k
Umk
=
1+
−
+
−
+
− ... −
+
Vk
Uk2
U2k
U3k
U3k
U4k
Umk
U(m+1)k
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k
(−1)mk U2k
1
(−1)mk Uk X
+
· Sk (m) −
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
1
(−1)mk U2k
Umk U2k
=
+
Sk (m)
1−1+
Vk
Uk U(m+1)k
U(m+1)k
−
=
k
(−1)mk Uk X
1
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
k
(−1)mk Uk X
1
Umk + (−1)mk Uk
Sk (m) −
.
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
The proof of the theorem: We give only the proof of (9), the proof of (10) is similar. Since
∞ X
1
1
−
Vn Vn+k . . . Vn+(m−1)k Vn+(m+1)k
Vn+k Vn+2k . . . Vn+mk Vn+(m+2)k
n=1
(26)
=
k
X
i=1
1
,
Vi Vi+k . . . Vi+(m−1)k Vi+(m+1)k
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using (22) and Lemma 2.1, then the left side of (26) can be written as
∞
X
Vn+mk Vn+(m+2)k − Vn Vn+(m+1)k
n=1
=
=
=
(27)
Vn Vn+k . . . Vn+(m+1)k Vn+(m+2)k
(−1)(m+1)k
U(m+1)k Sk (m) − (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
Uk
X
∞
1
(−1)(m+1)k
U(m+2)k
+ 1−
Uk
V
V
.
.
.
V
n+(m−1)k Vn+(m+1)k
n=1 n n+k
(−1)(m+1)k
U(m+1)k Sk (m) − (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
Uk
Umk +(−1)mk Uk
(−1)(m+1)k
U(m+2)k
Sk (m)
+ 1−
Uk
U(m+1)k
#
k
(−1)mk Uk X
1
−
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
(−1)k +(−1)mk −(−1)k V(m+1)k
Uk Sk (m)−(−1)k (p2 +4)Uk U(m+2)k Tk (m + 2)
U(m+1)k
−
k
(−1)mk Uk − (−1)k U(m+2)k X
1
.
U(m+1)k
VV
. . . Vi+mk
i=1 i i+k
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Considering the right sides of (26) and (27), we have
(−1)k + (−1)mk − (−1)k V(m+1)k
Uk Sk (m)
U(m+1)k
k X
(−1)mk Uk − (−1)k U(m+2)k Vi+(m+1)k + U(m+1)k Vi+mk
=
U(m+1)k Vi Vi+k . . . Vi+(m+1)k
i=1
+(−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
=
k
X
(−1)mk Vi+(m+1)k − (−1)k Vi+2(m+1)k
U(m+1)k Vi Vi+k . . . Vi+(m+1)k
i=1
Uk + (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2).
Hence
Sk (m)
=
1
1 + (−1)(m−1)k − V(m+1)k
"
k
X
(−1)(m−1)k Vi+(m+1)k − Vi+2(m+1)k
i=1
Vi Vi+k Vi+2k . . . Vi+(m+1)k
i
+(p2 + 4)U(m+1)k U(m+2)k Tk (m + 2) .
The proof of (9) is completed.
Acknowledgment. The authors would like to thank the anonymous referee for his/her valuable suggestions
that have significantly improved the presentation of this paper.
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1. Melham R. S., On some reciprocal sums of Brousseau: an alternative approach to that of Carlitz, The Fibonacci Quarterly,
41(1) (2003), 59–62.
2. Rabinowitz S., Algorithmic summation of reciprocals of products of Fibonacci numbers, The Fibonacci Quarterly, 37(2) (1999),
122–127.
3. Sun Z. H., Expansions and identities concerning Lucas sequences, The Fibonacci Quarterly, 44(2) (2006), 145–153.
4. Ye X. L. and Zhang Z. Z., On some formulas of the reciprocal sum and the alternating sum for generalized Fibonacci numbers,
Advanced Studies in Contemporary Mathematics, 10(2) (2005), 143–148.
Xiaoli Ye, Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P. R. China
Zhizheng Zhang, Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P. R. China,
e-mail: zhzhzhang-yang@163.com
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