IJAAMM
Int. J. Adv. Appl. Math. and Mech. 4(4) (2017) 1 – 4 (ISSN: 2347-2529)
Journal homepage: www.ijaamm.com
International Journal of Advances in Applied Mathematics and Mechanics
Some sum formulas for products of Pell and Pell-Lucas numbers
Research Article
Hasan GÖKBAŞ a, ∗ , Hasan KÖSE b
a Şemsi Tebrizi Girl Anatolian Religious Vocational High School, Konya, Turkey
b Science Faculty, Selcuk University, Konya, Turkey
Received 23 February 2017; accepted (in revised version) 12 April 2017
Abstract:
Let us define P and Q to be complex Pell and complex Pell-Lucas numbers. We have determined sum formulas for
squares of terms of these complex number sequences. We found some sum formulas for certain products of terms of
the Pell and Pell-Lucas sequences.
MSC:
11B39 • 11B99
Keywords: Complex Pell numbers • Complex Pell-Lucas numbers • Pell numbers • Pell-Lucas numbers
© 2017 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
Studies show that there has been an increasing interest in number sequences and their generalizations, such as Pell
and Pell-Lucas number sequences. These number sequences and generalizations have very important properties and
applications in almost every fields of science and art. Applications of these number sequences provide a wide area for
researchers. The features of these number sequences can be seen in [1–5].
The purpose of this paper is to derive some sum formulas for certain products of the terms of these number sequences. We also describe the complex Pell and complex Pell-Lucas numbers, like the complex Fibonacci and complex Lucas numbers [4]. We give formulae of sums for squares of the terms of these complex number sequences.
2. Preliminaries
The complex Pell and complex Pell-Lucas sequences P and Q are defined by recurrence relations
P 0 = i , P 1 = 2, P n = 2Pn + i Pn−1
f or
n≥2
Q0 = 4 − 2i , Q1 = 4 + 2i , Qn = 2Qn + i Qn−1 f or n ≥ 2
p
and i = −1. If start from n=0, then the complex Pell and complex Pell-Lucas sequence are given through
n
Pn
Qn
Pn
Qn
0
0
2
i
4-2i
1
1
2
2
4+2i
2
2
6
4+i
12+2i
3
5
14
10+2i
28+6i
4
12
34
24+5i
68+14i
5
29
82
58+12i
164+34i
···
···
···
···
···
∗ Corresponding author.
E-mail addresses: [email protected] (Hasan GÖKBAŞ), [email protected] (Hasan KÖSE).
Some sum formulas for products of Pell and Pell-Lucas numbers
2
The following sum formulas the Pell and Pell-Lucas numbers are well known [4, 5]:
n−1
X
P k2 =
k=1
n−1
X
k=1
n−1
X
Q k2 =
P n P n−1
2
Q 2n−1 − 2(−1)n − 4
2
P k P k+1 =
P 2n+1 − 2P n+1 P n − 1
4
Q k Q k+1 =
Q 2n − 2(−1)n − 8
2
k=1
n−1
X
k=1
n−1
X
P 2k+1 =
P 2n
2
Q 2k+1 =
Q 2n − 2
2
k=0
n−1
X
k=0
n−1
X
Q 2k =
k=0
Q 2n−1 + 2
2
3. Some sum formulas for Pell and Pell-Lucas sequences
Theorem 3.1.
If P n and Q n are nth Pell and Pell-Lucas numbers, we have
n
X
k=1
n
X
kP k P k−1 =
(2n + 1) (P 2n+1 − 2P n+1 P n − 1) − P 2n + 2n
8
kQ k Q k−1 =
2nQ 2n −Q 2n−1 + [4n + 2] (−1)n+1
4
k=1
Proof. Let A n =
n
X
n
P
k=1
P k P k−1 =
P 2n+1 −2P n+1 P n −1
, then
4
kP k P k−1 = P 1 P 0 + 2P 2 P 1 + 3P 3 P 2 + ... + nP n P n−1
k=1
=
n
X
k=1
P k P k−1 +
n
X
k=2
P k P k−1 +
n
X
P k P k−1 + ... +
k=3
n
X
P k P k−1
k=n
= A n + (A n − A 1 ) + (A n − A 2 ) + ... + (A n − A n−1 )
¶ n−1 µ
¶
µ
n−1
X
X P 2i +1 − 2P i +1 P i − 1
P 2n+1 − 2P n+1 P n − 1
−
= n An −
Ai = n
4
4
i =1
i =1
µ
¶
P 2n+1 − 2P n+1 P n − 1
P 2n − 2 P 2n+1 − 2P n+1 P n − 1 2n − 2
=n
−
+
+
4
8
8
8
(2n + 1) (P 2n+1 − 2P n+1 P n − 1) − P 2n + 2n
=
8
So, the proof is completed. Accordingly, the value of
n
P
k=1
kQ k Q k−1 is computed.
Theorem 3.2.
If P n and Qn are nth complex Pell and complex Pell-Lucas numbers, we have
n
X
¶
4P n+1 P n − P n P n−1
=
+ i (P 2n+1 − 2P n+1 P n − 1)
2
k=1
µ
¶
n
X
¡
¢
4Q 2n+1 −Q 2n−1 − 20 + 10(−1)n
Qk2 =
+ i 2Q 2n − 4(−1)n
2
k=1
P k2
µ
Hasan GÖKBAŞ, Hasan KÖSE / Int. J. Adv. Appl. Math. and Mech. 4(4) (2017) 1 – 4
3
Proof.
n
X
k=1
P k2 =
n
X
(2P k + i P k−1 )2
k=1
n
X
P k2 + i 2
=4
k=1
n
X
k=1
2
P k−1
+ 4i
n
X
P k P k−1
k=1
¶ µ
¶
µ
¶
µ
P n+1 P n
P n P n−1
P 2n+1 − 2P n+1 P n − 1
=4
−
+ 4i
2
2
4
¶
µ
4P n+1 P n − P n P n−1
+ i (P 2n+1 − 2P n+1 P n − 1)
=
2
So, the proof is completed. Accordingly, the value of
n
P
k=1
Qk2 is computed.
Theorem 3.3.
If P n and Qn are nth complex Pell and complex Pell-Lucas numbers, we have
n
X
¸
4n (4P n+1 P n − P n P n−1 ) − 4P 2n+1 + 8P n+1 P n + P 2n−1 − 2P n P n−1 + 3
8
k=1
·
¸
2n (P 2n+1 − 2P n+1 P n ) + P 2n+1 − P 2n − 2P n+1 P n − 1
+i
2
h
i
 2n(4Q 2n+1 −Q 2n−1 )−4Q 2n +Q 2n−2 −20n−18 + i [2nQ 2n −Q 2n−1 + 4n + 2] , if n is odd
n
X
4
2
i
h
kQk =
 2n(4Q 2n+1 −Q 2n−1 )−4Q 2n +Q 2n−2 +20n+2 + i [2nQ 2n −Q 2n−1 − 4n − 2] , otherwise
k=1
4
kP k2 =
Proof. Let B n =
n
X
k=1
·
n
P
k=1
P k2 =
³
4P n+1 P n −P n P n−1
2
´
+ i (P 2n+1 − 2P n+1 P n − 1),then
kP k2 = P 12 + 2P 22 + 3P 32 + ... + nP n2
=
n
X
k=1
P k2 +
n
X
k=2
P k2 +
n
X
k=3
P k2 + ... +
n
X
k=n
P k2
= B n + (B n − B 1 ) + (B n − B 2 ) + ... + (B n − B n−1 )
¶
¸
·µ
n−1
X
1
= nB n −
B j = n 2P n+1 P n − P n P n−1 + i (P 2n+1 − 2P n+1 P n − 1)
2
j =1
·µ
¶
¸
n−1
X
¡
¢
1
−
2P j +1 P j − P j P j −1 + i P 2 j +1 − 2P j +1 P j − 1
2
j =1
¸
·
4n (4P n+1 P n − P n P n−1 ) − 4P 2n+1 + 8P n+1 P n + P 2n−1 − 2P n P n−1 + 3
=
8
·
¸
2n (P 2n+1 − 2P n+1 P n ) + P 2n+1 − P 2n − 2P n+1 P n − 1
+i
2
So, the proof is completed. Accordingly, the value of
n
P
k=1
kQk2 is computed.
Corollary 3.1.
P n and Qn are nt h complex Pell and complex Pell-Lucas numbers, we have formulas for
n
P
k=1
kP k2 and
n
P
k=1
kQk2 .
Some sum formulas for products of Pell and Pell-Lucas numbers
4
We can derive a formula for
n
P
k=1
n
X
k=1
(n + 1 − k)P k2 = (n + 1)
(n + 1 − k) P k2 and
n
X
k=1
P k2 −
n
X
k=1
n
P
k=1
(n + 1 − k) Qk2 .
kP k2
¶
¸
4P n+1 P n − P n P n−1
+ i (P 2n+1 − 2P n+1 P n − 1)
2
¸
·
4n (4P n+1 P n − P n P n−1 ) − 4P 2n+1 + 8P n+1 P n + P 2n−1 − 2P n P n−1 + 3
−
8
¸
·
2n (P 2n+1 − 2P n+1 P n ) + P 2n+1 − P 2n − 2P n+1 P n − 1
−i
2
¶
µ
¶
µ
P 2n+1 − 2P n+1 P n + P 2n − 2n − 1
8P n+1 P n − 2P n P n−1 + 4P 2n+1 − P 2n−1 − 3
+i
=
8
2
·µ
= (n + 1)
Using the same technique, we can be show that
n
X
k=1
(n + 1 − k)Qk2
h
i
 8Q 2n+1 +4Q 2n −2Q 2n−1 −Q 2n−2 −40n−42 + i [2Q 2n +Q 2n−1 + 2] ,
4
i
= h 8Q
2n+1 +4Q 2n −2Q 2n−1 −Q 2n−2 −40n−22

+ i [2Q 2n +Q 2n−1 − 2] ,
4
if n is odd
otherwise
References
[1] R. Melham, Sums Involving Fibonacci and Pell Numbers, Portugalie Math. 56(3) (1999) 309 - 317.
[2] A. F. Horadam, Pell Identities, The Fibonacci Quart. 9(3) (1971) 245 - 252.
[3] S. Vajda, Fibonacci and Lucas Numbers and the Golden Section: Theory and Applications, Ellis Horwood Ltd.,
England, 1989.
[4] T. Koshy, Fibonacci and Lucas Numbers with Applications, A Wiley-Interscience Publication, New York, 2001.
[5] T. Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
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