A NEW RELATION BETWEEN THE FIBONACCI AND LUCAS SEQUENCES FOR 7-STEP

  Bulletin of Mathematics Vol. 08, No. 02 (2016), pp. 159–168.

  

A NEW RELATION BETWEEN THE

FIBONACCI AND LUCAS SEQUENCES FOR

  7

• -STEP

  

M. R. Fadhilla, S. Gemawati, I. Hasbiyati

and M. D. H. Gamal

  Abstract.

  This article discusses a relation of Fibonacci and Lucas sequences having seven order. This formula is used to determine the kth term of Lucas 7 step. This paper uses the method of Natividad and Policarpio [General Mathe- matics Notes, 7 (2013), 82–87] and applies mathematical induction to prove the formula. The result indicates that every variable of Lucas 7 step forms Fibonacci 7 step. It can be concluded that to determine kth term of Lucas 7 step can be ob- tained by summing up multiplication of the first seventh term of Lucas 7 step and corresponding Fibonacci 7 step.

1 INTRODUCTION

  

The Fibonacci sequence (F n ) is recursive sequence obtained by summing up

previous two numbers. This sequence has experience expansion by many

researchers. The expansion is done not limited to the formula of finding

Fibonacci sequence, but also is developed to Fibonacci like sequence.

  The Fibonacci sequence is not the only sequence obtained by summing

up previous two numbers. Similar to Fibonacci is called Lucas sequence

(L

  n

  ). The difference of both sequences lay in initial conditions, that is Received 30-11-2016, Accepted 23-12-2016. M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas Step

  7 F = 0, F

  1 = 1 and L = 2, L 1 = 1. Many researchers study relationship

  

of these sequences. Benyamin and Quinn [2] find identity of the relation

of Fibonacci and Lucas sequences. Parindeni and Gemawati [5] find the

formula for Lucas like sequence of fourth step and fifth step. Azarian [1] also

finds the relation of Fibonacci and Lucas sequences for binomial quantifying

identity.

  Tony and Post [4] study the identity of the relation of Fibonacci and

Lucas sequences having order n equivalent to the relation of recursive of the

following

  (n) (n) (n)

  − F = 2F F ,

  k k

• 1 k−n (n) (n) (n)

  − L = 2L L ,

  k k

• 1 k−n (n) (n) (n) (n) (n)

  L = F + 2F + · · · + (n − 1)F + nF .

  k k k−1 k−n+2 k−n+1

2 FIBONACCI n STEP AND LUCAS n STEP

  

The following gives by definition of Fibonacci n step and Lucas n step and

some examples of it.

  (n)

  Definition 2.1 The Fibonacci n step (F ) defined as formula linear re-

  k

  cursive is having order n > 1:

  (n) (n) (n) (n)

  

F ,

= F + F + · · · + F (1)

  k k

• 1 k−1 k−n+1

  (n) (n) where F = 1, F = 0, k = −n + 2, · · · , 0. k

  1 (n)

  Definition 2.2 The Lucas n step (L ) defined as formula linear recursive

  k

  is having order n > 1:

  (n) (n) (n) (n)

  

L ,

= L + L + ... + L (2)

  k k−1 k−2 k−n (n) (n) where L = n and L = −1 for k = −n + 1, · · · , −1. k

  

Based on equation (1) and equation (2), the example for Fibonacci n step

and Lucas n step for 2 ≤ n ≤ 7 can be showed in Table 1 and Table 2.

  

3. A NEW RELATION THE FIBONACCI AND LUCAS FOR

SEVENTH ORDER

Furthermore we give the expansion process of the relation of Fibonacci 7

step and Lucas 7 step. To find the relation it starts by analysing some

equations of Lucas 7 step obtained from equation (2) for 8 ≤ k ≤ 15 and

  1

  5

  1

  3

  7

  15

  31 57 113 223 439 863 1695 3333 6553

  6

  3

  26

  7

  15

  31 63 120 239 475 943 1871 3711 7359

  7

  1

  3

  7

  51 99 191 367 708 1365 2631 5071

  15

  31 63 127 247 493 983 1959 3903 7775 Parindeni and Gemawati [5] find a new formula to determine the kth term of Lucas 4 step (L

  3

  3

  4

  7

  11

  18

  29

  47 76 123 199 322 521

  1

  7

  3

  7

  11

  21

  39 71 131 241 443 815 1499 2757

  4

  1

  3

  15

  (4) n

  2

  3

  (4)

  2

  ∗ k−2

  ∗ k−3

  ∗ k−4

  )L

  (4)

  ∗ k−2

  ∗ k−3

  ∗ k−3

  ∗ k−4

  ∗ k−5

  )L

  (4)

  4 .

  In this paper, we study a new relationship of Fibonacci and Lucas

sequences for seventh order so that the formula is applicable to determine

k th term of Lucas sequence for seventh order.

  )L

  ∗ k−2

  ). The formula obtained uses the method of

Natividad dan Policarpio [3]. By [5, Theorem 1] for every Lucas sequence

L

  3

  (4)

  1

  ,L

  (4)

  2

  ,L

  (4)

  and L

  1

  (4)

  4

  applies L

  (4) k

  =(M

  ∗ k−2

  )L

  (4)

  1

  13

  M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas

  1

  8

  13

  

21

  34

  55 89 144 233 377

  3

  1

  2

  3

  4

  7

  13

  24

  

44

81 149 274 504 927 1705

  4

  1

  5

  2

  2

  7

  7 Step

Table 1: The Fibonacci n step sequences

k

  1

  2

  3

  4

  5

  6

  

8

  1

  9

  10

  11

  12

  13

  14 n

  2

  1

  1

  4

  12

  3

  2

  4

  8

  16

  32 64 127 253 504 1004 2000 3984

Table 2: The Lucas n step sequences

k

  1

  2

  4

  1

  5

  6

  7

  8

  9

  10

  11

  1

  7

  8

  8

  15

  29 56 108 208 401 773 1490 2872

  5

  1

  1

  2

  4

  16

  32 63 125 248 492 976 1936 3840

  31 61 120 236 465 912 1793 3525

  6

  1

  1

  2

  4

  8

  16

• (M

• + M

• (M
• M
• M
• (M
• M
• M
• M

M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas Step

  7 n = 7 as follows:

  

  (7) (7) (7) (7) (7) (7) (7) (7)

  L = L + L + L + L + L + L + L 

  8

  1

  2

  3

  4

  5

  6

  7

   

  (7) (7) (7) (7) (7) (7) (7) (7)

    L = L + 2L + 2L + 2L + 2L + 2L + 2L 

  9

  1

  2

  3

  4

  5

  6

  7

   

  (7) (7) (7) (7) (7) (7) (7) (7)

   L  = 2L + 3L + 4L + 4L + 4L + 4L + 4L

  10

  1

  2

  3

  4

  5

  6 7 

   

  (7) (7) (7) (7) (7) (7) (7) (7)

   L = 4L + 6L + 7L + 8L + 8L + 8L + 8L

  11

  1

  2

  3

  4

  5

  6

  7 (7) (7) (7) (7) (7) (7) (7) (7)

  

L = 8L + 12L + 14L + 15L + 16L + 16L + 16L 

  12

  1

  2

  3

  4

  5

  6

  7

   

  (7) (7) (7) (7) (7) (7) (7) (7) 

   L = 16L + 24L + 28L + 30L + 31L + 32L + 32L 

  13

  1

  2

  3

  4

  5

  6

  7

   

  (7) (7) (7) (7) (7) (7) (7) (7)

    L = 32L + 48L + 56L + 60L + 62L + 63L + 64L 

  14

  1

  2

  3

  4

  5

  6

  7

   

  (7) (7) (7) (7) (7) (7) (7) (7)

   L

= 64L + 96L + 112L + 120L + 124L + 126L + 127L

  15

  1

  2

  3

  4

  5

  6

  7

  (3)

  (7) (7) (7) (7) (7) (7)

  , L , L , L , L Based on equation (3), coefficients of L , L and

  1

  2

  3

  4

  5

  6 (7) L for 8th until kth are presented in Table 3.

  7 (7) (7) (7) (7) (7) (7) (7)

  Table 3: Coefficient arrangements of L ,L ,L ,L ,L ,L ,L

  1

  2

  3

  4

  5

  6

  7

  (7) k L Coefficients k (7) (7) (7) (7) (7) (7) (7)

  L L L L L L L

  1

  2

  3

  4

  5

  6

  7 (7) L

  8

  1

  1

  1

  1

  1

  1

  1

  8 (7) L

  9

  1

  2

  2

  2

  2

  2

  2

  9 (7)

  10 L

  2

  3

  4

  4

  4

  4

  4

  10 (7) L

  11

  4

  6

  7

  8

  8

  8

  8

  11 (7)

  12 L

  8

  12

  14

  15

  16

  16

  16

  12 (7)

  13 L

  16

  24

  28

  30

  31

  32

  32

  13 (7) L

  14

  32

  48

  56

  60

  62

  63

  64

  13 (7)

  15 L

  64 96 112 120 124 126 127

  13

.. .. .. .. .. .. .. .. ..

  

. . . . . . . . .

_{P P P P P} (7) (7) 8 (7) 9 (7) 10 (7) 11 (7) 12 (7) (7)

k L F F F F F F F

k k i k i i k i i k i i k i i k i k _{−7 =7 − =7 − =7 − =7 − =7 − −8}

  (7) (7) (7)

  From Table 3 it can be showed that the coefficients of L , L , L ,

  1

  2

  3 (7) (7) (7) (7) L , L , L and L are Fibonacci 7 step as showed in Table 1 for n = 7.

  4

  5

  6

  7 (7)

  The coefficient of L for the 8th term to 15th is the first eight term of the

  1 (7) (7) M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas Step

  7

  (6)

  

the coefficient of L for 9th is summing up the 8th and 9th the coefficient

  2 (7) (7)

  (7)

  

L , and L for 10th is summing up 9th and dan 10th the coefficient L ,

  1

  2

  1 (7) (7) (7)

  so that the coefficient of kth term of L can be expresed as F + F

  2 k−7 k−8

  P

  8 (7) or F . i

  =7 k−i (7)

  Hereinafter the coefficient of L for 10th is summing up 8th, 9th and

  3 (7) (7)

  

10th from the coefficent L and the coefficient of L for 11th is summing

  1

  3 (7)

  

up the 9th, 10th and 11th term coeficient of L , so that the coefficient of

  1 (7) (7) (7) (7) P 9 (6)

  

k th term of L can be expressed as F + F + F or F . The

  i

  3 k−7 k−8 k−9 =7 k−i (7) (7)

  

pattern continues until L . While the pattern of coefficient L for 8th to

  6

  7 (7) k th are corresponding F . k−6

  (7)

  From visible above observation that coefficient of L is corresponding

  1 (7) (7) P (7) (7) P (7) (7) P (7) (7)

  8

  9

  10

  

to F , L to F , L to F , L to F , L to

  i i i 2 =7 3 =7 4 =7

  5 k−7 k−i k−i k−i

  P (6) (7) P (7) (7) (7)

  11

  12 F , L to F and L to F . So that we obtain the i i =7 6 =7

  7 k−i k−i k−6

  relation of Fibonacci 7 step and Lucas 7 step as follows:

  (7) (7) (7) (7) (7) (7) (7)

  Theorem 2.1 For every L , L , L , L , L , L , L are Lucas 7

  1

  2

  3

  4

  5

  6

  7

  step, the formula to determine the k-th term of Lucas n step is

  8

  9

  10 X

  X X

  (7) (7) (7) (7) (7) (7) (7) (7) (7)

  L F L

  F L F L =F +

• +

  L

  k

  1

  2

  3

  4 k−7 k−i k−i k−i i i i

  =7 =7 =7

  11

  12 X

  X

  (7) (7) (7) (7) (7) (7)

  

L F L L ,

• 5
• F
• F (4)

  6

  7 k−i k−i k−6 i i

  =7 =7 (7) (7) (7)

  

where L is the k-th term of Lucas 7 step,, L are the first, L is the

  k

  1

  2 (7) (7) (7) (7) (7)

  

second, L is third, L is fourth, L is fifth L is sixth, L is seventh

  3

  4

  5

  6

  7 (7) (7) (7) (7) (7) (7) (7)

  

and F , F ,F ,F ,F ,F , F are corresponding Fibonacci

  k−6 k−7 k−8 k−9 k−10 k−11 k−12 7 step.

  Proof.

  The theorem is proved using the strong mathematical induction

for k ≥ 8 and k is some positive integers. Assume that k = 8 is bases, so

that we get

  (7) 8−i

  (7)

  (7) 8−i

• 8

• 9

  L

  F

• 10

  F

  X

  2

  (7)

  L

  (7) 8−i

  F

  1

  X

  X

  (7)

  (7)

  =F

  8

  (7)

  7 Step L

  M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas

  L

  (7)

  4

  3

1 L

  i =7

• (
• (

• + F

• (1 + 0)L
• (1 + 0 + 0)L
• (1 + 0 + 0 + 0)L
• (1 + 0 + 0 + 0 + 0)L
• 1 + 0 + 0 + 0 + 0 + 0)L
• (1)L
• L
• L
• L
• L
• L
• L

• 8
• 9
• 10

  5

  (7)

  4

  X

  i =8

  F

  (7) m−i

  L

  (7)

  X

  (7) m−i

  i =8

  F

  (7) m−i

  L

  (7)

  6

  (7) m−7

  L

  (7)

  L

  i =8

  F

  L

  =F

  (7) m−8

  L

  (7)

  1

  X

  i =8

  F

  (7) m−i

  (7)

  , (7) P

  2

  X

  i =8

  F

  (7) m−i

  L

  (7)

  3

  X

  7

  (m − 2) : L

  , (6) P (m − 1) : L

  X

  (7)

  4

  X

  i =9

  F

  (7) m−i

  L

  (7)

  5

  i =9

  (7) m−i

  F

  (7) m−i

  L

  (7)

  6

  (7)

  L

  (7)

  , (8)

  L

  F

  (7) m−2

  L

  =F

  (8) m−9

  L

  (8)

  1

  X

  i =9

  F

  (7) m−i

  (7)

  i =9

  2

  X

  i =9

  F

  (7) m−i

  L

  (7)

  3

  X

  (7) m−1

  (7)

  7

  8

  (7)

  4

  (7)

  5

  (7)

  6

  (7)

  7 L (7)

  =L

  (7)

  (7)

  1

  (7)

  2

  (7)

  3

  (7)

  4

  (7)

  3

  2

  i =7

  (7) 8−i

  i =7

  11 X i =7

  F

  (7) 8−i

  )L

  (7)

  5

  12 X i =7

  F

  )L

  (7)

  (7)

  6

  (7)

  2 L (7)

  7 L (7)

  8

  =(1)L

  (7)

  1

  5

  (7)

  6

  (7)

  (7) m−i

  L

  (7)

  4

  X

  F

  (7) m−i

  L

  5

  (7)

  X

  i =7

  

F

  (7) m−i

  L

  (7)

  6

  (7) m−6

  L

  F

  i =7

  X

  3

  7 .

  (5) Equation (5) is true from equation (2) for n = 7 and k = 8. Next assume that the statement is true for k = 8, 9, 10, · · · , m − 6, m − 5, m − 4, m − 3, m − 2, m − 1, m with m is some positive integers.

  P (m) : L

  (7) m =F (7) m−7

  L

  (7)

  1

  X

  i =7

  F

  (7) m−i

  L

  (7)

  2

  X

  i =7

  F

  (7) m−i

  L

  (7)

• 11
• 12
• F

  i =7

• 9
• 10
• 11
• 12
• 13
• F
• 10
• 11
• 12
• 13
• 14
• F

M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas Step

  7

  11

  12 X

  X

  (7) (7) (7) (7) (7) (7) (7)

• P (m − 3) : L =F L F L F L

  m−3 m−10 1 m−i 2 m−i

  3 i i =10 =10

  13

  14

  15 X

  X X

  (7) (7) (7) (7) (7) (7)

  F L F L F L

  m−i 4 m−i 5 m−i

  6 i i i =10 =10 =10

  (6) (7)

• F L ,

  (9)

  m−9

  7

  12

  13 X

  X

  (7) (7) (7) (7) (7) (7) (7)

• P (m − 4) : L =F + L F L F L

  1

  2

  3 m−4 m−11 m−i m−i i i

  =11 =11

  14

  15

  16 X

  X X

  (7) (7) (7) (7) (7) (7)

• F L F L F L

  4

  5

  6 m−i m−i m−i i i i

  =11 =11 =11 (6) (7)

• F L , (10)

  7 m−10

  13

  14 X

  X

  (7) (7) (7) (7) (7) (7) (7)

  P F L F L

• L (m − 5) : L =F
• m−5 m−12

  1

  2

  3 m−i m−i i i

  =12 =12

  15

  16

  17 X

  X X

  (7) (7) (7) (7) (7) (7)

  F L F L F L

• + +

  +

  4

  5

  6 m−i m−i m−i i i i

  =12 =12 =12 (6) (7)

  L ,

• F

  (11)

  m−11

  7

  14

  15 X

  X

  (7) (7) (7) (7) (7) (7) (7)

  (m − 6) : L + + P =F L F

  2 F L

  1

  3 m−6 m−13 m−i m−i i i

  =13 =13

  16

  17

  18 X

  X X

  (7) (7) (7) (7) (7) (7)

• F L F L F L

  m−i 4 m−i 5 m−i

  6 i i i =13 =13 =13

  (6) (7)

• F L . (12)

  7 m−12

  Next P (m + 1) is proved true if P (m), P (m − 1), P (m − 2), P (m − 3), P (m − 4), P (m − 5), P (m − 6), · · · , 10, 9, 8 are true, that is

  7

  8 X

  X

  (7) (7) (7) (7) (7) (7) (7)

  =F + + P (m + 1) : L L F L F L

  m

• 1

  1

  2

  3 m−6 m−i m−i i i

  =6 =6

  9

  10

  11 X

  X X

  (7) (7) (7) (7) (7) (7)

• F L F L F L

  4

  5

  6 m−i m−i m−i i i i

  =6 =6 =6 (7) (7)

• L
• L
• L
• L
• L
• 8
• 9
• 10
• 11
• 12
• F
• L
• L
• L
• L
• L
• L
• 8

• +

  9
• 10
• 1

• 11
• 12
• F
• F
• 9
• 10
• 11
• 12
• 13
• F
• F
• 10
• 11
• 12
• 13
• 14
• F
• F
• 11
• 12
• 13
• 14
• 15
• F
• F

  (7)

  7

  (8) m−9

  L

  (8)

  1

  X

  i =9

  F

  (7) m−i

  L

  (7)

  2

  i =9

  X

  F

  (7) m−i

  L

  (7)

  3

  X

  i =9

  F

  (7) m−i

  L

  (7)

  4

  X

  L

  (7) m−7

  6

  L

  2

  X

  i =8

  F

  (7) m−i

  L

  (7)

  3

  X

  i =8

  F

  (7) m−i

  (7)

  (7)

  4

  X

  i =8

  F

  (7) m−i

  L

  (7)

  5

  X

  i =8

  F

  (7) m−i

  L

  i =9

  (7) m−i

  F

  5

  i =10

  F

  (7) m−i

  L

  (7)

  4

  X

  i =10

  F

  (7) m−i

  L

  (7)

  X

  3

  i =10

  F

  (7) m−i

  L

  (7)

  6

  (7) m−9

  L

  (7)

  7

  (7) m−11

  L

  (7)

  X

  (7)

  L

  (7) m−10

  (7)

  5

  X

  i =9

  F

  (7) m−i

  L

  (7)

  6

  (7) m−8

  L

  (7)

  7

  L

  L

  (7)

  1

  X

  i =10

  F

  (7) m−i

  L

  (7)

  2

  X

  i =10

  F

  (7) m−i

  (7)

  (7) m−i

  L

  (7)

  X

  i =7

  F

  (7) m−i

  L

  (7)

  3

  X

  i =7

  F

  (7) m−i

  L

  4

  (7)

  X

  i =7

  F

  (7) m−i

  L

  (7)

  5

  X

  i =7

  F

  (7) m−i

  L

  (7)

  2

  L

  (7) m−6

  , so that equation (6) becomes L

  M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas

  7 Step

To prove equation ( 13), both sides of equation (6) are summing up with

L

  (7) m−1

  , L

  (7) m−2

  , L

  (7) m−3

  , L

  (7) m−4

  , L

  (7) m−5

  

and L

  (7) m−6

  (7) m + L (7) m−1

  (7) m−i

  (7) m−2

  (7) m−3

  (7) m−4

  (7) m−5

  (7) m−6

  = F

  (7) m−7

  L

  (7)

  1

  X

  i =7

  F

  6

  L

  F

  F

  L

  (7)

  4

  X

  i =7

  F

  (7) m−i

  L

  (7)

  5

  X

  i =7

  (7) m−i

  F

  L

  (7)

  6

  (7) m−6

  L

  (7)

  7

  (7) m−8

  L

  (7)

  1

  X

  i =8

  (7) m−i

  i =7

  (7)

  1

  7

  (7) m−1

  (7) m−2

  (7) m−3

  (7) m−4

  (7) m−5

  (7) m−6

  . (14) Hereinafter substituting of equations (7), (8), (9), (10), (11) and (12) into equation (14) we obtain L

  (7) m

  =F

  (7) m−7

  L

  (7)

  X

  X

  i =7

  F

  (7) m−i

  L

  (7)

  2

  X

  i =7

  F

  (7) m−i

  L

  (7)

  3

  1

• 12
• 13
• 14
• 15
• 16
• F
• F
• 13
• 14
• 15
• 16
• 17
• F
• F
• 14
• 15
• 16

• 17
• 18
• F
• 13
• 14
• 13
• 1
• 14
• 15

• +

  13
• 14
• 15
• 16
• 13
• 14
• 15
• 16
• 17
• 13
• 14
• 15
• 16
• 17
• 18

• +

  12
• 7

• +

  8
• 9
• 1

  X

  (7) m−1

  F

  i =8

  X

  (7) m−i

  i =7

  F

  X

  4

  (7)

  L

  (7) m−i

  F

  i =9

  F

  (7) m−i

  X

  i =10

  F

  (7) m−i

  X

  i =11

  F

  (7) m−i

  L

  (7)

  5

  X

  i =7

  F

  i =10

  (7) m−i

  X

  X

  i =8

  F

  (7) m−1

  L

  (7)

  2

  X

  i =7

  F

  (7) m−i

  X

  i =8

  F

  (7) m−1

  i =9

  F

  (7) m−i

  i =9

  X

  (7) m−1

  F

  i =8

  X

  F

  F

  i =7

  X

  3

  (7)

  L

  (7) m−i

  (7) m−i

  i =8

  X

  i =6

  i =6

  F

  (7) m−i

  L

  (7)

  2

  X

  i =6

  F

  (7) m−i

  L

  (7)

  3

  X

  F

  1

  (7)

  L

  (7) (7)

  L

  (7)

  F

  11 X

  L

  (7) m−i

  (7)

  F

  10 X

  4

  (7)

  L

  X

  (7)

  F

  i =12

  (7) m−1

  X

  i =9

  F

  (7) m−i

  X

  i =10

  F

  (7) m−i

  X

  i =11

  F

  (7) m−i

  X

  F

  L

  L

  (7) m−6

  =F

  (7) m

  (16) From (1) for n = 7, it is obtained that L

  7

  (7)

  (7) m−i

  (7) m−i

  F

  i =6

  X

  6

  (7)

  L

  X

  F

  (7) m−i

  L

  (7)

  6

  (6) m−10

  L

  (7)

  7

  (7) m−12

  L

  (7)

  1

  X

  i =12

  F

  (7) m−i

  (7)

  (7) m−i

  X

  4

  (7)

  L

  (7) m−i

  F

  i =12

  3

  2

  (7)

  L

  (7) m−i

  F

  i =12

  X

  L

  F

  i =12

  (7)

  M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas

  7 Step

  X

  i =11

  F

  (7) m−i

  L

  (7)

  2

  X

  i =11

  F

  (7) m−i

  L

  3

  i =11

  i =11

  X

  5

  (7)

  L

  (7) m−i

  F

  X

  X

  4

  (7)

  L

  (7) m−i

  F

  i =11

  X

  F

  i =7

  L

  L

  (7)

  4

  X

  i =13

  F

  (7) m−i

  L

  (7)

  5

  X

  i =13

  F

  (7) m−i

  (7)

  F

  13 X i =7

  X

  1

  (7)

  L

  (7) m−i

  F

  =

  6

  (7) m

  (15) Next the equation (15) can be made as L

  7 .

  (7)

  L

  (6) m−12

  (7) m−i

  i =13

  (7) m−i

  (7) m−13

  L

  (7)

  5

  X

  i =12

  F

  (7) m−i

  L

  (7)

  6

  (6) m−11

  L

  (7)

  7

  L

  X

  X

  3

  (7)

  L

  (7) m−i

  F

  i =13

  (7)

  (7)

  2

  (7) m−i

  F

  i =13

  X

  1

  (7) . M. R. Fadhilla, S. Gemawati, I. Hasbiyati & M. D. H. Gamal – The Formula Lucas

  7 Step The last statement is an analogical correctness from (13) so that P (m + 1) is correct. So the proof follows.

  

REFERENCES

  

1. M. K. Azarian, Identities involving Lucas or Fibonacci and Lucas numbers

as binomial sums, International Journal of Contemporary Mathematical Sciences, 7 (2012), 2221–2227.

  2. A. T. Benjamin dan J.J Quinn, Recounting Fibonacci and Lucas identities, The College Mathematics Journal, 30 (1999), 359–366.

  3. L. R. Natividad and P. B. Policarpio, A novel formula in solving tribonacci- like sequence, General Mathematics Notes, N.7, 2013, pp.82–87.

  

4. T. D. Noe dan J. V. Post, Primes in Fibonacci n step and Lucas n step

sequences, Journal of Integer Sequences, 8 (2005), 1–12.

  

5. R. Parindeni dan S. Gemawati, Formula for Lucas Like Sequence of Fourth

Step and Fifth Step, International Mathematical Forum, 12 (2017), 103– 110.

  Mushbar Rabby Fadhilla : Master’s Student, Department of Mathematics, Fac-

ulty of Mathematics and Natural Sciences, University of Riau, Bina Widya Campus,

  Pekanbaru 28293, Indonesia.

  E-mail: mushbar.fadhilla@gmail.com Sri Gemawati

  : Department of Mathematics, Faculty of Mathematics and Natural

Sciences, University of Riau, Bina Widya Campus, Pekanbaru 28293, Indonesia.

  E-mail: gemawati.sri@gmail.com Ihda Hasbiyati

  : Department of Mathematics, Faculty of Mathematics and Natural

Sciences, University of Riau, Bina Widya Campus, Pekanbaru 28293, Indonesia.

  E-mail: ihdahasbiyati26@yahoo.com M. D. H Gamal

  : Department of Mathematics, Faculty of Mathematics and Natural

Sciences, University of Riau, Bina Widya Campus, Pekanbaru 28293, Indonesia.

  E-mail: mdhgamal@unri.ac.id