Int. J. Adv. Appl. Math. and Mech. 3(1) (2015) 98 – 101 (ISSN: 2347-2529) Journal homepage:www.ijaamm.com
International Journal of Advances in Applied Mathematics and Mechanics
On some properties of (k,h)-Pell sequence and (k,h)-Pell-Lucass
sequence
Research Article
Seyyed Hossein Jafari-Petroudia,∗, Behzad Pirouzb
aDepartment of Mathematics, Payame Noor University, P. O. Box 1935-3697, Tehran, Iran
bDepartment of Mathematics, Azad University of Karaj, Karaj, Iran
Received 22 May 2015; accepted (in revised version) 15 August 2015
Abstract: In this note we first define (k,h)-Pell sequence and (k,h)-Pell-Lucas sequence. Then we derive some formulas fornt h term and sum of the first n terms of these sequences. Finally other properties of these sequences are represented. MSC: 15A15• 15A60
Keywords:Pell-Lucass sequence• Recursion relation• Sum
© 2015 The Author(s). This is an open access article under the CC BY-NC-ND license(https://creativecommons.org/licenses/by-nc-nd/3.0/).
1. Introduction
In [1] Bueno studied (k,h)-Jacobsthal sequence of the form
Tn=kTn−1+2hTn−2.
He found a formula ofnt hterm and sum of the first n terms of this sequence. In this note we first define (k,h)-Pell sequence and (k,h)-Pell-Lucas sequence. Then we derive some formulas fornt h term and sum of the first n terms of these sequences. Finally other properties of these sequences are represented. For more information about (k ,h)-Jacobsthal sequence, Fibonacci sequence and some generalizations of this sequence see [1] - [8].
Pell sequence {Pn} has the recursive relation
Pn=2Pn−1+Pn−2,
whereP0=0,P1=1. Now we define a generalization of this sequence which we call it (k,h)-Pell sequence and denote
it byΦn. This sequence has the recursive relation
Φn=2kΦn−1+hΦn−2, (1)
whereΦ0=0 andΦ1=2kandk,h∈Zandk2+h>0. Also we define (k,h)-Pell-Lucas sequence {Λn} which has the
recursive relation
Λn=2kΛn−1+hΛn−2, (2)
whereΛ0=2 andΛ1=2k. It is known that
n−1 X
k=0
xk=1+x+x2+ · · · +xn−1=x
n
−1
x−1 , (3)
n−1 X
k=1
kxk=(n−1)x
n
−nxn−1+1
(x−1)2 . (4)
∗ Corresponding author.
Jafari-Petroudi / Int. J. Adv. Appl. Math. and Mech. 3(1) (2015) 98 – 101 99
2. Main result
Theorem 2.1.
LetΦnbe as in(1)then we have
Φn=2k
p
¡
αn−βn¢
,
whereα=k+ p
k2+h,β=k− p
k2+h and p=α−β.
Proof. The recursive relation (1) has the characteristic equation
r2−2kr−h=0.
The roots of this equation areα=k+ p
k2+h,β=k− p
k2+h. Also we haveα+β=2k,α−β= p
k2+h=pand
αβ= −h. So the solution of the recursion relation (1) is
Φn=c1αn+c2βn. (5)
If we use the initial valuesΦ0=0 andΦ1=2kwe get a linear system with two equationsc1+c2=0 andc1α+c2β=2k.
This linear system has the solutionc1=2k
p andc2=
−2k
p . By substituting these values in (5) we get
Φn=2k
p
¡
αn−βn¢
.
Theorem 2.2.
LetΛnbe as in(2)then we have
Λn=αn+βn,
whereα=k+ p
k2+h,β=k− p
k2+h andΛ0=2,Λ1=2k.
Proof. The proof is similar toTheorem 2.1.
Theorem 2.3.
LetΦnbe as in(1)then we have
n−1 X
m=0Φm
=Φn+hΦn−1−2k
2k+h−1 .
Proof. ByTheorem 2.1we have
n−1 X
m=0
Φm=2k
p
n−1 X
m=0 ¡
αm−βm¢=2k
p
"n−1 X
m=0
αm−
n−1 X
m=0
βm #
According to (3) we get
n−1 X
m=0
Φm=2k
p
·1 −αn
1−α −
1−βn
1−β ¸
=2k
p
·(1
−αn)(1−β)−(1−βn)(1−α)
(1−α)(1−β)
¸
.
After some calculations we get
n−1 X
m=0
Φm=2k
p
µ(
α−β)−(αn−βn)+αβ(αn−1−βn−1)
1−(α+β)+αβ
¶
=2k
p
"pΦ1 2k −
pΦn
2k +(−h) pΦn−1
2k
1−2k−h
#
=Φ1−Φn−hΦn−1
1−2k−h =
Φn+hΦn−1−2k
2k+h−1 .
So we have
n−1 X
m=0Φm
=Φn+hΦn−1−2k
100 On some properties of (k,h)-Pell sequence and (k,h)-Pell-Lucass sequence
Theorem 2.4.
LetΦnbe as in(1)then we have
n−1 X
m=0ΦmΦm−1
=4k
2
p2
·2k
−Λ3+hΛ2n−1−h3Λ2n−3
(−h)3+(−h)−(−h)Λ2 +
2k h
µ1 −(−h)n
1+h ¶¸
.
Proof. ByTheorem 2.1we have
n−1 X
m=0
ΦmΦm−1= µ2k
p
¶2n−1 X
m=0 ¡
αm−βm¢ ¡αm−1−βm−1¢
=4k 2 p2 "n−1 X m=0
α2m−1+
n−1 X
m=0
β2m−1−(α+β)
n−1 X
m=0
(αβ)m−1
# =4k 2 p2 "n−1 X m=0
α−1(α2)m+
n−1 X
m=0
β−1(β2)m−(α+β) αβ
n−1 X
m=0
(αβ)m
#
According to (3) we get
n−1 X
m=0
ΦmΦm−1=4k
2
p2
·
(1
α)
1−α2n
1−α2 +(
1
β)
1−β2n
1−β2 −(
2k
−h)
1−(αβ)n
1−αβ ¸ . =4k 2 p2 ·1 −α2n α−α3 +
1−β2n β−β3 +
2k h
µ1 −(−h)n
1+h
¶¸
.
After some calculations we get
n−1 X
m=0
ΦmΦm−1=4k
2
p2
·(
α+β)−(α3+β3)−αβ(α2n−1+β2n−1)+α3β3(α2n−3+β2n−3)
(αβ)3+αβ−(αβ)(α2+β2)
¸ +4k 2 p2 ·2k h µ1 −(−h)n
1+h
¶¸
.
So byTheorem 2.1andTheorem 2.2we conclude that
n−1 X
m=0
ΦmΦm−1=4k
2
p2
·2k
−Λ3+hΛ2n−1−h3Λ2n−3
(−h)3+(−h)−(−h)Λ2 +
2k h
µ1 −(−h)n
1+h
¶¸
.
Theorem 2.5.
LetΦnbe as in(1)then we have
n−1 X
m=0 Φ2m=
4k2 p2
·
h2Λ2n−2−Λ2n−Λ2+2
h2−Λ2+1 +2
(−h)n−1
h+1 ¸
.
Proof. ByTheorem 2.1we have
n−1 X m=0Φ 2 m= n−1 X m=0 ·2k
p (α
m
−βm) ¸2 =4k 2 p2 n−1 X m=0 ¡
α2m+β2m−2(αβ)m¢
=4k 2 p2 "n−1 X m=0
(α2)m+
n−1 X
m=0
(β2)m−2
n−1 X
m=0
(αβ)m
#
.
According to (3) we get
n−1 X
m=0 Φ2m=
4k2 p2
· α2n−1
α2−1 +
β2n−1 β2−1 −2
(αβ)n
−1 αβ−1
¸
=4k
2
p2 ·
α2β2(α2n−2+β2n−2)−(α2n+β2n)−(α2+β2)+2
(αβ)2−(α2+β2)+1
¸ +4k
2
p2 ·
2(−h)
n
−1
h+1 ¸
.
So byTheorem 2.1andTheorem 2.2we deduce that
n−1 X
m=0 Φ2m=
4k2 p2
·
h2Λ2n−2−Λ2n−Λ2+2
h2−Λ2+1 +2
(−h)n−1
h+1 ¸
Jafari-Petroudi / Int. J. Adv. Appl. Math. and Mech. 3(1) (2015) 98 – 101 101
Theorem 2.6.
LetΛnbe as in(2)then we have
n−1 X
m=0Λm
=Λn+hΛn−1+2k−2
2k+h−1 .
Proof. The proof is similar toTheorem 2.3.
Theorem 2.7.
LetΛnbe as in(2)then we have
n−1 X
m=0
ΛmΛm−1=2k−Λ3+hΛ2n−1−h
3 Λ2n−3
−h(1+Λ2+h2) + 2k
h
·(
−h)n−1
h+1 ¸
.
Proof. The proof is similar toTheorem 2.4.
Theorem 2.8.
LetΛnbe as in(2)then we have
n−1 X
m=0 Λ2m=
2−Λ2+h2Λ2n−2−Λ2n
1−Λ2+h2 +2
1−(−h)n
1+h .
Proof. The proof is similar toTheorem 2.5.
Acknowledgments
Authors are very grateful to reviewers for their significant explanation for enhancement of this paper.
References
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