Title Science and Technology Indonesia e-ISSN:2580-4391 p-ISSN:2580-4405 Vol. 4, No. 2, April 2019 Research Paper Counting the sum of cubes for Lucas and Gibonacci Numbers Wamiliana1*, Suharsono1, P. E. Kristanto2 1Dept. of Mathematics, Universitas Lampung, Jl. Soemantri Brojonegoro No. 1, Bandarlampung, Indonesia 2SMP Xaverius 1, Jl. Sultan Hasanuddin No. 20, Telukbetung, Bandarlampung, Indonesia *Corresponding author: wamiliana.1963@fmipa.unila.ac.id Abstract Lucas and Gibonacci numbers are two sequences of numbers derived from a welknown numbers, Fibonacci numbers. The di�erence between Lucas and Fibonacci numbers only lies on the first and second elements. The first element in Lucas numbers is 2 and the second is 1, and nth element, n ≥ 3 determined by similar pa�ern as in the Fibonacci numbers, i.e : Ln = Ln-1 + Ln-2. Gibonacci numbers G_0,G_1,G_2,G_3. . . ; G_n=G_(n-1)+G_(n-2) are generalized of Fibonacci numbers, and those numbers are nonnegative integers. If G_0=1 and G_1=1, then the numbers are the wellknown Fibonacci numbers, and if G_0=2 and G_1=1, the numbers are Lucas numbers. Thus, the di�erence of those three sequences of numbers only lies on the first and second of the elements in the sequences. For Fibonacci numbers there are quite a lot identities already explored, including the sum of cubes, but there have no discussions yet about the sum of cubes for Lucas and Gibonacci numbers. In this study the sum of cubes of Lucas and Gibonacci numbers will be discussed and showed that the sum of cubes for Lucas numbers is ∑ni=0 (Li)3= Ln(Ln+1) 2+5(−1)nLn−1 + 19 2 and for Gibonacci numbers is ∑ n i=0 (Gi) 3= [Gn(Gn+1) 2+(G1−G0)3−3G02G1+4G03−(−1)n(G12−G1G0−G02)Gn−1] 2 Keywords Fibonacci numbers, Lucas numbers, Gibonacci numbers, identities, sum of cubes Received: 11 Februari 2019, Accepted: 8 April 2019 https://doi.org/10.26554/sti.2019.4.2.31-35 1. INTRODUCTION Fibonacci number was originally stated by Leonardo Pisano or Fibonacci in his book, Liber Abaci in 1202. In his book he stated a problem about the growth of pairs rabbits. A young pair of rabbits (male and female) is placed on an island. A pair of rabbits does not breed until they are two months old. After they are two month old, each pair of rabbits produces another pair each month. Assuming that there are no rabbits ever die then the total number of pair of rabbits form a sequence 1,1,2,3,5,8,13,...and this sequence of numbers is known as Fibonacci numbers. Lucas number was introduced by François Edouard Anatole Lucas in 1877 by setting the �rst number is 2 and the second is 1, and the other numbers are following the pattern as on Fibonacci numbers so that it form a sequence of numbers 2, 1, 3, 4, 7, 11, 18, 29, 47,. . . Gibonacci numbers G0,G1,G2,G3,...;Gn = Gn−1 + Gn−2 are generalized of Fibonacci numbers, and those numbers are nonnegative integers. If G0 = 1 and G1 = 1,then the numbers are the welknown Fibonacci numbers, and if G0 = 2 and G1 = 1, the numbers are Lucas numbers. Thus, the di�erence of those three sequence of numbers only lies on the �rst and second of the elements in the sequences. There are many identities already developed that related among Gibonacci, Fibonacci, and Lucas numbers. Posamentier et al. (2007) showed that Fibonacci, Lucas, and Gibonacci num- bers related to Pascal triangle, art and golden ratio. There are some researchers already investigated identities for Fibonacci Lucas and Gibonacci numbers such as Carlitz et al. (1972a,b,c,d), and Benjamin et al. (1999, 2000, 2009, 2003). Krishna (1980), in- vestigated generalized Fibonacci sequence and William (1972) discussed how to get Fibonacci numbers from Pascal’s triangle. Dunlap (2003) investigated about the golden ratio and Gener- alized Fibonacci numbers, including the application in biology and crystallography. Benjamin et al. (2003) proved more than one hundred Fibonacci identities by combinatorial arguments and Benjamin et al. (2009) proposed the identity for �nding the sum of cubes for Fibonacci numbers. Frontczak (2018) discussed about the relationship of sum of powers for Fibonacci and Lucas numbers for sum of �rst, second, third, and fourth powers; and Arangala et al. (2016) investigated the relationship of the sum of power for Fibonacci and Lucas numbers for the power of square, fourth, sixth, and eighth. However, until recently there have no discussion about the sum of cubes for Lucas numbers where in the formula just contains the Lucas numbers itself, or the sum of cubes for Gibonacci numbers where in the formula just contains the Gibonacci numbers itself. Therefore, in this study we will discuss about the sum of cubes for Gibonacci numbers as an https://doi.org/10.26554/sti.2019.4.2.31-35 Wamiliana et. al. Science and Technology Indonesia, 4 (2019) 31-35 extension identity of Koshy (2001), and Benjamin et al. (1999). In Section 2 and 3 we will present the results from Koshy (2001), Benjamin et al. (1999), and Dunlap (2003), and in Section 4 we will present the results. 2. Some Identities for Lucas and Gibonacci Numbers 2.1 Some Identities for Lucas Number The Identity 1, 2, and 3 are proposed by Koshy (2001), and the Identity 4 is proposed by Benjamin et al. (1999). Identity 1 L0 + L1 + L2 + ... + Ln = Ln+2 − L1 Proof. Using the de�nition of Lucas number we get Ln−2 = Ln − Ln−1 then, L0 = L2 − L1 L1 = L3 − L2 L2 = L4 − L3 L3 = L5 − L4 . . . Ln−1 = Ln+1 − Ln Take the sum on both sides simultaneously we get : L0 + L1 + L2 + L3+...+Ln = Ln+2 − L1 Identity 2 L0 + L1 + L2 + L3 + ... + L2n = L2n+1 − L1 Proof. By de�nition, Ln−1 = Ln − Ln−2 then, L0 = L0 L2 = L3 − L1 L4 = L5 − L3 L6 = L7 − L5 . . . L2n−2 = L2n−1 − L2n−3 L2n = L2n+1 − L2n−1 Take the sum on both sides simultaneously we get : L0 + L1 + L2 + L3+...+L2n = L2n+1 − L1 + L0 = L2n+1 − 1 + 2 = L2n+1 + 1 = L2n+1L1 Identity 3 L1 + L3 + L5+...+L2n−1 = L2n − L0 Proof. By de�nition, Ln−1 = Ln − Ln−2 then, L1 = L2 − L0 L3 = L4 − L2 L5 = L6 − L4 . . . L2n−3 = L2n−2 − L2n−4 L2n−1 = L2n − L2n−2 Take the sum on both sides simultaneously we get : L1 + L3 + L5+...+L2n−1 = L2n − L0 Identity 4 Ln+1Ln−1 = Ln2 − (−1)n⋅5 Proof. Using mathematical induction, and we leaven the basic induction and just do the induction step. Asumme that the statement is true for n = k , Lk+1Lk−1 = Lk2 − (−1)k⋅5 For n = k + 1, then : Lk+12 = Lk+1⋅Lk+1 Lk+12 = (Lk+2 − Lk)Lk+1 Lk+12 = Lk+2⋅Lk+1 - Lk⋅Lk+1 Lk+12 = Lk+2 − (Lk + Lk−1)-Lk⋅Lk+1 Lk+12 = Lk+2Lk + Lk−1+Lk+2⋅Lk−1 − Lk⋅Lk+1 Lk+12 = Lk+2Lk + (Lk−1 + Lk)Lk−1⋅LkLk+1 Lk+12 = Lk+2Lk+Lk+1Lk−1 + Lk⋅Lk−1 − Lk⋅Lk+1 Lk+12 = Lk+2Lk + Lk+1⋅Lk−1 + Lk(Lk+1 − Lk−1) Lk+12 = Lk+2Lk + Lk+1Lk−1 + Lk(−Lk) Lk+12 = Lk+2Lk + Lk+1Lk−1−Lk2 Since, Lk−1Lk+2=−Lk2−(−1) k⋅5, then Lk+12=Lk+2⋅Lk−(−1) k⋅5 Lk+12=Lk+2⋅Lk+(−1) k + 1⋅5 Lk+2⋅Lk=Lk+12+(−1) k + 1⋅5 2.2 Some Identities for Gibonacci Number The Identity 5,6, and 7 are proposed by Dunlap (2003), Identity 8 is proposed by Benjamin et al. (2009), while Identity 9 derived from De�nition 8. Those Identity will be used in the process to determine the sum of cubes for Gibonacci numbers. Identity 5 G0 + G1 + G2+...+Gn = Gn+2 − G1 Proof. Using the de�nition of Gibonacci numbers, we get Gn−2 = Gn − Gn−1 then, G0 = G2 − G1 G1 = G3 − G2 G2 = G4 − G3 G3 = G5 − G4 . . . Gn−1 = Gn+1 − Gn Gn = Gn+2 − Gn+1 Take the sum of both sides simultaneously we get: G0 + G1 + G2+...+Gn = Gn+2 − G1 Identity 6 G0 + G1 + G2 + G3+...+G2n = G2n+1 + G0 − G1 © 2019 The Authors. Page 32 of 36 Wamiliana et. al. Science and Technology Indonesia, 4 (2019) 31-35 Proof. Again, using the de�nition of Gibonacci numbers, we get Gn−1 = Gn − Gn−2, G0 = G0 G2 = G3 − G1 G4 = G5 − G3 G6 = G7 − G5 . . . G2n−2 = G2n−1 − G2n−3 G2n = G2n+1 − G2n−1 Take the sum of both sides simultaneously we get : G0 + G1 + G2 + G3+...+G2n = G2n+1 + G0 − G1 Identity 7 G1 + G3 + G5+...+G2n−1 = G2n − G0 Proof. Gn−1 = Gn − Gn−2 then, G1 = G2 − G0 G3 = G4 − G2 G5 = G6 − G4 . . . G2n−3 = G2n−2 − G2n−4 G2n−1 = G2n − G2n−2 Take the sum on both sides simultaneously we get : G1 + G3 + G5+...+G2n−1 = G2n − G0 Identity 8 Gn+1Gn−1 = Gn2 − (−1)n⋅{G12 − G1G0 − G02} Proof. By using mathematical induction. For n = k + 1, then : G2⋅G0 = Gn2 − (−1)n⋅{G12 − G1G0 − G02} (G1 + G0)G0 = G12 − G12 − G1G0 − G02 G1G0 + G02 = G1G0G02 Assume that the statement is true for n = k, then Gk+12 = Gk+1⋅Gk+1 Gk+12 = (Gk+2 − Gk)Gk+1 Gk+12 = Gk+2⋅Gk+1 - Gk⋅Gk+1 Gk+12 = Gk+2 − (Gk + Gk−1)-Gk⋅Gk+1 Gk+12 = Gk+2Gk + Gk−1+Lk+2⋅Gk−1 − Gk⋅Gk+1 Gk+12 = Gk+2Gk + (Gk−1 + Gk)Gk−1⋅GkGk+1 Gk+12 = Gk+2Gk+Gk+1Gk−1 + Gk⋅Gk−1 − Gk⋅Gk+1 Gk+12 = Gk+2Gk + Gk+1⋅Gk−1 + Gk(Gk+1 − Gk−1) Gk+12 = Gk+2Gk + Gk+1Gk−1 + Gk(−Gk) Gk+12 = Gk+2Gk + Gk+1Gk−1−Gk2 Because, Gn+1Gn−1 = Gn2 − (−1)n⋅{G12 − G1G0 − G02} then Gk+12=Gk+2⋅Gk−(−1) k⋅{G12 − G1G0 − G02} Gk+12=Gk+2⋅Gk+(−1) k + 1⋅{G12 − G1G0 − G02} We get Gk+2⋅Gk=Gk+12+(−1) k + 1⋅{G12 − G1G0 − G02} The following identity was derived from Identity 8 Identity 9 2Gn2=Gn+12−GnGn−1 − (−1)n(G12 − G1G0 − G02) Proof. Gn2=Gn⋅Gn Gn2=Gn⋅(Gn+1 − Gn−1) Gn2=Gn⋅Gn+1−Gn⋅Gn−1 Gn2=(Gn+1 − Gn−1)⋅Gn+1 − Gn⋅Gn−1 Gn2=Gn+12−Gn+1⋅Gn−1 − Gn⋅Gn−1 Using Identity 4 Gn+1Gn−1 = Gn2 − (−1)n⋅{G12 − G1G0 − G02} Then: Gn2=Gn+12{Gn2 − (−1)n⋅(G12 − G1G0 − G02)}⋅Gn−1 − Gn⋅Gn−1 2Gn2=Gn+12−GnGn−1 − (−1)n(G12 − G1G0 − G02) 3. Sum of Cubic for Lucas and Gibonacci Numbers Based on the De�nition above, we can derive an De�nition for �nding the sum of cubes for Lucas and Gibonacci number as follow: Identity 10 The sum of cubic of Lucas numbers is ∑ni=0 (Li) 3= Ln(Ln+1) 2+5(−1)nLn−1 + 19 2 Proof. Using Identity 5: 2Ln2=Ln+12−LnLn−1 − (−1)n⋅5, then 2L03 = 2L03 2L13 = 2L1L12 2L13 = L1(L22L1L0 + (−1)15) 2L13 = L1L22 − L12⋅−5L1 2L23 = 2L2L22 2L23 = L2(L32L2L1 + (−1)15) 2L23 = L2L12 − L22⋅−5L1 2L33 = 2L3L32 2L33 = L3(L42L3L2 + (−1)15) 2L33 = L3L42 − L32⋅−5L1 2L43 = 2L4L42 2L43 = L4(L52L4L3 + (−1)15) 2L43 = L4L32 − L42⋅−5L1 2L53 = 2L5L52 2L53 = L5(L62L5L4 + (−1)15) 2L53 = L5L42 − L52⋅−5L1 2L63 = 2L6L62 2L63 = L6(L72L6L5 + (−1)15) 2L63 = L6L52 − L62⋅−5L1 . . . 2Ln3 = 2LnLn2 2Ln3 = Ln(Ln+12LnLn−1 + (−1)15) 2Ln3 = LnLn−12 − Ln2⋅−5L1 Take the sum on both sides simultaneously we get : 2(L03 + L13 + L23 + L33+...+Ln3) 2(L03 − L12L0 − 5(L1 − L2 + L3 − L4 + L5 − L6 + ... − (−1)nLn)LnL2n+1 LnL2n+1 − 5(L1 − L2 + L3 − L4 + L5 − L6+...−(−1) nLn2(L03 − L12L0 © 2019 The Authors. Page 33 of 36 Wamiliana et. al. Science and Technology Indonesia, 4 (2019) 31-35 Next, we need to �nd the sum of L1 − L2 + L3 − L4 + L5 − L6+...−(−1)nLn By using Identity 2 and 3, we get: −L2 − L4 − L6+...L2n = −(L2n+1 + L1 − L0) (1) −L1 − L3 − L5+...L2n−1 = −(L2n − L0) (2) From equation (3), adding both sides with L2n − L0, equation (3) becomes: −L2 − L4 − L6+...L2n + L2n − L0 = −(L2n+1 + L1 − L0) − (L2n − L0) (3) By using equation (4), equation (5) becomes : −L2 − L4 − L6 + ...L2n + (−L1 − L3 − L5+...L2n−1) = −(L2n+1 + L1 − L0) − L2n − L0 L1 − L2 + L3 − L4 + L5 − L6 + ... + L2n−1 − L2n = −L2n+1 + L1 − L0 − L2n − L0 = −L2n+1 + L1 − L2n = −L2n−1 − L2n (4) Adding L2n+1 to both sides of equation (5), we get : L1 − L2 + L3 − L4 + L5 − L6 + ...L2n−1 − L2n + L2n+1 = −L2n−1 − L1 + L2n+1 L1 − L2 + L3 − L4 + L5 − L6 + ...L2n−1 − L2n + L2n+1 = L2n+1 − L2n−1 − L1 Tℎus, L1 − L2 + L3 − L4 + L5 − L6 + ...L2n−1 − L2n + L2n+1 = L2n − L1 (5) By equation (6) and (7), we can conclude that : L1 −L2 +L3 −L4 +L5 −L6 + ...−(−1)nLn = −(−1)nLn−1 − L1 (6) Therefore, by using equation (8), equation (3) becomes: 2(L03 + L13 + L23 + L33+...+Ln3) = LnL2n+1 − 5[−(−1) nLn−1 − L1]2(L03 − L12L0 = L2n+1 + 5(−1) nLn−1 + 5L1 + 2L03 − L12L0 (7) By using de�nition that L0 = 2 and L1 = 1, then: 2(L03 + L13 + L23 + L33+...+Ln3) = LnL2n+1 − 5[−(−1) nLn−1 + 5 + 16 − 2 = LnL2n+1 − 5[−(−1) nLn−1 + 19 Tℎeref ore ∶ n ∑ i=0 (Li)3 = Ln(Ln+1)2 + 5(−1)nLn−1 + 19 2 Example : Given Lucas sequence : 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, . . . , then 2(L03 + L13 + L23 + L33 + ... + L103) = L10(L11)2 + 5(−1)10L9 + 19 2 = 123(199)2 + 5(−1)1076 + 19 2 = 123(39601) + 5(−1)1076 + 19 2 = 4870923 + 380 + 19 2 = 2435661 By using direct summation we get : 2(L03 + L13 + L23 + L33 + ... + L103) = 23 + 13 + 33 + 43 + 73 + 113 + 183 + 293 + 473 + 763 + 1233 = 8 + 1 + 27 + 64 + 343 + 1331 + 5832 + 24389 + 103823+ 438976 + 1860867 = 2435661 Identity 11 ∑ni=0 (Gi)3= Gn(Gn+1) 2 + G1 − G03 − 3G02G14G03−(−1)n(G12 − G1G0 − G02)Gn−1 2 Proof. By using Identity 9: 2Gn2=Gn+12−GnGn−1 − (−1)n(G12 − G1G0 − G02) 2G03 = 2G03 2G13 = 2G1G12 2G13 = G1(G2 − G1G0 − (−1)n(G12 − G1G0 − G02)) 2G13 = G1G22 − G22⋅G0 + (G12 − G1G0 − G02)G1 2G23 = 2G2G22 2G23 = G2(G3 − G2G1 − (−1)n(G12 − G1G0 − G02)) 2G23 = G2G32 − G32⋅G0 + (G12 − G1G0 − G02)G2 2G33 = 2G3G32 2G33 = G3(G4 − G3G2 − (−1)n(G12 − G1G0 − G02) 2G33 = G3G42 − G42⋅G0 + (G12 − G1G0 − G02)G3 2G43 = 2G4G42 2G43 = G4(G5 − G3G3 − (−1)n(G12 − G1G0 − G02)) 2G43 = G4G52 − G52⋅G0 + (G12 − G1G0 − G02)G4 2G53 = 2G5G52 2G53 = G5(G6 − G4G2 − (−1)n(G12 − G1G0 − G02)) 2G53 = G5G62 − G52⋅G0 + (G12 − G1G0 − G02)G5 2Gn3 = 2GnGn2 2Gn3 = Gn(Gn+1 − GnGn−1 − (−1)n(G12 − G1G0 − G02)) 2Gn3 = GnGn+12 − Gn2⋅Gn−1(−1)n(G12 − G1G0 − G02)Gn © 2019 The Authors. Page 34 of 36 Wamiliana et. al. Science and Technology Indonesia, 4 (2019) 31-35 By taking the sum of both sides simultaneously we get: 2(G03 + G13 + G23 + G33+...+Gn3) = 2G03 − G12.G0 +(G12 − G1G0 − G02)(G1 − G2 + G3 − G4 + G5 − G6 +... − (−1)nGn) + GnGn+12 (8) Next, we need to �nd (G1 − G2 + G3 − G4 + G5 − G6+...−(−1)nGn) + GnGn+12 By using Identity 2, we get : −G2 − G4 − G6...−G2n = −(G2n+1 − G1) (9) and Identity 7: G1 + G3 + G5...+G2n−1 = G2n − G0 (10) From equation (11), add both sides of the equation with G2n − G0 then equation (12) becomes : −G2−G4−G6...−G2n +G2n −G0 = (G2n−1−G1)+G2n −G0 (11) By equation (12), equation(13) will be : −G2 − G4 − G6... − G2n + G2n + G1 + G3 + G5... + G2n−1 = (G2n−1 − G1) + G2n − G0 G1 − G2 + G3 − G4 + G5 + G6 + ... + G2n−1 − G2n = G2n−1 + G2n + G1 − G0 (12) From equation (13), add both sides withG2n+1, we get: G1 − G2 + G3 − G4 + G5 + G6+...+G2n−1 − G2n + G2n+1 = G2n−1 + G1 − G0 + G2n+1 = G2n+1 − G2n−1 + G1 − G0 By using the de�nition we get G2n = G2n+1 − G2n−1 thus, G1 − G2 + G3 − G4 + G5 + G6 + ... + G2n−1 − G2n + G2n+1 = G2n + G1 − G0 (13) From equations (14) and (15), we can conclude that : G1 − G2 + G3 − G4 + G5 − G6 + ... − (−1)nGn = −(−1)nGn−1+ + G1 − G0 (14) By using equation (16), equation (17) become: 2G30 + G 3 1 + G 3 2 + G 3 3 + ... + G 3 n = 2G30 − G 2 1 .G0 + (G 2 1 − G1G0 − G 2 0) [−(−1)nGn−1 + G1 − G0] + GnG2n+1 = GnG2n+1 + 2G 3 0 − G 2 1G0− (−1)n + G21 − G1G0 − G 2 0Gn−1 + G1 − G0 = GnG2n+1 + 2G 3 0 − G 2 1 .G0 + (G 2 1 − G1G0 − G 2 0) [−(−1)nGn−1 + (G21 − G1G0 − G 2 0)(+G1 − G0) = GnG2n+1 + 2G 3 0 − G 2 1G0 + (G 2 1 − G1G0 − G 2 0) (+G1 − G0) − (−1)n(G21 − G1G0 − G 2 0)Gn−1 = GnG2n+1 + 3G 3 0 − G 3 13G 2 1G0− (−1)n(G21 − G1G0 − G 2 0)Gn−1 = GnG2n+1 + (G1 − G0) 3 + 4G30 − 3G 2 0G1− (−1)n(G21 − G1G0 − G 2 0)Gn−1 (15) Therefore: G03 + G13 + G23+...+Gn3= [Gn(Gn+1)2+(G1−G0)3−3G02G1+4G03−(−1)n(G12−G1G0−G02)Gn−1] 2 Example : For G0 = 3 and G1 = 2 , the Gibonacci sequence is 3, 2, 5, 7, 12, 19, 31, 50, 81, 131, 212,. . . G03 + G13 + G23+...+G93= [G9(G10)2+(G1−G0)3−3G02G1+4G03−(−1)9(G12−G1G0−G02)G8] 2 G03 + G13 + G23+...+G93= [131(212)2+(2−3)3−3322+443−(−1)9(22−23−3)281] 2 G03 + G13 + G23+...+G93= [(131)(44944)−1−54−108+(−11)81] 2 G03 + G13 + G23+...+G93= [5887664−1λ54+108−891] 2 =2943413 By using direct summation we get : G03 + G13 + G23+...+G93= 33 + 23 + 53 + 73 + 123 + 193 + 313 + 503 + 813 + 1313 = 9 + 8 + 125 + 343 + 1728 + 6859 + 29791 + 125000 + 531441 + 2248091 = 2943413 4. CONCLUSIONS From the discussion above we can conclude that the sum of cubic for Lucas number is L03 + L13 + L23 + L33+...+Ln3= ∑ni=0 (Li)3= Ln(Ln+1) 2+5(−1)nLn−1 + 19 2 and for Gibonacci numbers is G03 + G13 + G23 + G33+...+Gn3= ∑ni=0 (Gi) 3= [Gn(Gn+1) 2+(G1−G0)3−3G02G1+4G03−(−1)n(G12−G1G0−G02)Gn−1] 2 REFERENCES Arangala, C., B. Hrovat, and J. Kelner (2016). Sum of Powers of Fibonacci and Lucas Numbers. The Minnesota Journal of Undergraduate Mathematics, 2(1); 1–12 Benjamin, A. T, T. A. Carnes, and B. Cloitre (2009). Recounting the Sums of Cubes of Fibonacci Numbers. Congressus Numer- antium, Proceedings of the Eleventh International Conference © 2019 The Authors. Page 35 of 36 Wamiliana et. al. Science and Technology Indonesia, 4 (2019) 31-35 on Fibonacci Numbers and their Applications. William Webb, ed, 194; 45–51 Benjamin, A. T, and J. J. Quinn (1999). Recounting Fibonacci and Lucas Identities. College Mathematics Journal, 30(5); 359 – 366 Benjamin, A. T, and J. J. Quinn (2003). Proofs That Really Count : The Art of Mathematical Association of America. Washington DC Benjamin, A. T, J. J. Quinn, and F. E. Su (2000). Phased Tilings and Generalized Fibonacci. The Fibonacci Quarterly, 38(3); 282–288 Carlitz, L., R. Scoville, and V. E. Hoggatt (1972a). Fibonacci Representations. Fibonacci Quarterly, 10(1); 1 – 28 Carlitz, L., R. Scoville, and V. E. Hoggatt (1972b). Fibonacci Representations. Fibonacci Quarterly, 10(1); 29 – 42 Carlitz, L., R. Scoville, and V. E. Hoggatt (1972c). Fibonacci Representations of Higher Order. Fibonacci Quarterly, 10(1); 71 – 80 Carlitz, L., R. Scoville, and V. E. Hoggatt (1972d). Fibonacci Representations of Higher Order. Fibonacci Quarterly, 10(1); 43 – 70 Dunlap, R. A. (2003). The Golden Ratio and Fibonacci Numbers. World Scienti�c Press, Singapore Frontczak, R. (2018). Sums of powers of Fibonacci and Lucas num- bers: A new bottom-up approach. Notes on Number Theory and Discrete Mathematics, 24(2); 94 –103 Koshy, T. (2001). Fibonacci and Lucas Numbers with Application. John Wiley & Sons Inc., New York Krishna, H. V. (1980). Identities of A Generalized Fibonacci Sequence. The Fibonacci Association, 18; 65 – 66 Posamentier, A. S, and I. Lehmann (2007). The Fabulous Fibonacci Numbers. Prometheus Books, New York William, H. C. (1972). Fibonacci Numbers Obtained from Pascal’s Triangle with Generalization. Fibonacci Quarterly, 10(4); 405 – 412 © 2019 The Authors. Page 36 of 36 INTRODUCTION Some Identities for Lucas and Gibonacci Numbers Some Identities for Lucas Number Some Identities for Gibonacci Number Sum of Cubic for Lucas and Gibonacci Numbers CONCLUSIONS