Sums of fourth powers of Fibonacci and Lucas numbers

Transcription

1 Sums of fourth powers of Fibonacci Lucas numbers arxiv: v1 [math.nt] 28 May 2017 Kunle Adegoke Department of Physics Engineering Physics, Obafemi Awolowo University, Ile-Ife, Nigeria Abstract We obtain closed-form expressions for all sums of the form n F mk 4 n k 4 their alternating versions, where F i L i denote Fibonacci Lucas numbers respectively. Our results complement those of Melham who studied the alternating sums. 1 Introduction The Fibonacci numbers, F n, Lucas numbers, L n, are defined, for n Z, as usual, through the recurrence relationsf n = F n 1 +F n 2,F 0 = 0, F 1 = 1 L n = L n 1 +L n 2, L 0 = 2, L 1 = 1, with F n = ( 1) n 1 F n L n = ( 1) n L n. About two decades ago, motivated by the results of Clary Hemenway [1] who obtained factored closed-form expressions for sums of the form n F mk 3, Melham [2] obtained factored closed-form expressions for alternating sums of the form n ( 1)k 1 F 4 mk. AMS Classification Numbers : 11B37, 11B39 adegoke00@gmail.com, kunle.adegoke@yex.com 1

2 Since no evaluations were reported in the Melham paper for the non alternating sums, we have attempted to fill that gap in this paper. Our main results are the following, valid for integers m n, with m not equal to zero: F mk 4 = n+m(n+m +4( 1) mn 1 ) k 4 = n+m(n+m +4( 1) mn ) +6n+3 +6n 5. We also re-derived the alternating sums, in slightly different but equivalent forms to the results contained in [2]: ( 1) k 1 F 4 mk = F { } mnf mn+m ( 1) n 1 n n+m +( 1) n(m 1) 4 5 k=(1+( 1) n )/2 ( 1) k 1 k 4 = ( 1)n 1 5F mn F mn+m { n n+m +( 1) nm 4 }, valid for all integers m n. 2 Required identities preliminary results 2.1 Telescoping summation identities The following telescoping summation identities are special cases of the more general identities proved in [3]. Lemma 2.1. If f(k) is a real sequence m n are positive integers, then [f(mk +m) f(mk)] = f(mn+m) f(m). Lemma 2.2. If f(k) is a real sequence m n are positive integers, then ( 1) k 1 [f(mk +m)+f(mk)] = ( 1) n 1 f(mn+m)+f(m). 2

3 2.2 First-order Lucas summation identities Lemma 2.3. If m n are integers, then F m ( 1) mk 1 k = ( 1) mn 1 F mn n+m. Proof. Setting v = m u = 2mk in the identity gives F u+v ( 1) v F u v = F v L u, (2.1) k+m k m = F m k, m even, (2.2) k+m +k m = F m k, m odd. (2.3) Using identity (2.2) in Lemma 2.1 with f(k) = F 2k m, it is established that F m k = F m+2mn F m on account of identity (2.1). = F m+mn+mn F m+mn mn = F mn n+m, m even, (2.4) Similarly, using identity (2.3) in Lemma 2.2 with f(k) = F 2k m, we have F m ( 1) k 1 k = ( 1) n 1 F m+2mn +F m = ( 1) n 1 (F m+mn+mn ( 1) n F m+mn mn ) = ( 1) n 1 (F m+mn+mn ( 1) mn F m+mn mn ), since m is odd = ( 1) n 1 F mn n+m, m odd. Identities (2.4) (2.5) combine to give Lemma (2.3). Lemma 2.4. If m n are integers, then ( 1) k(m 1) k = ( 1) n(m 1) n+m. 3 (2.5)

4 Proof. Setting v = m u = 2mk in the identity L u+v +( 1) v L u v = L v L u, (2.6) gives k+m k m = k, m odd, (2.7) k+m +k m = k, m even. (2.8) Using (2.7) in Lemma 2.1 with f(k) = L 2k m, we have k = +2mn, m odd. (2.9) Similarly, using (2.8) in Lemma 2.2 with f(k) = L 2k m, we have ( 1) k 1 k = ( 1) n 1 +2mn + m even. (2.10) Identities (2.9) (2.10) combine to give Lemma (2.4). 3 Main results 3.1 Non alternating sums Theorem 3.1. If m is a non-zero integer n is any integer, then F mk 4 = n+m(n+m +4( 1) mn 1 ) Proof. By squaring the identity making use of the identity +6n+3. 5F u 2 = L 2u ( 1) u 2, u Z, (3.1) L v 2 = L 2v +( 1) v 2, v Z, (3.2) finally setting u = mk, it is established that F mk 4 = L 4mk +( 1) mk 1 4k +6. (3.3) 4

5 By summing both sides of identity (3.3), using Lemma 2.3 to sum each of the first two terms on the right h side, we have F mk 4 = (nn+2m +4( 1) mn 1 F mn n+m ) +6n, (3.4) Using the identity (2.1) we can write n n+2m = F 4mn+2m = n+m n+m (3.5) F mn n+m = (n+m ( 1) mn F m ) = n+m +( 1) mn 1. (3.6) Substituting (3.5) (3.6) into (3.4) proves Theorem 3.1. Corollary 3.2. If n is an integer, then F 4 k = F 2n+1 L n 1 L n+2 +6n+3. Proof. From Theorem 3.1 we have F 4 k = F 2n+1 (L 2n+1 +4( 1) n 1 )+6n+3. (3.7) From identity (2.6) with u = n+2 v = n 1 we have L 2n+1 +4( 1) n 1 = L 2n+1 +( 1) n 1 L 3 = L n 1 L n+2, (3.8) the result follows. Theorem 3.3. If m is a non-zero integer n is any integer, then k 4 = n+m(n+m +4( 1) mn ) +6n 5. 5

6 Proof. The theorem is proved by summing both sides of the following identity, k 4 = L 4mk ( 1) mk 1 4k +6, (3.9) applying Lemma 2.3 to sum each of the first two terms on the right h side. Identity (3.9) is obtained by squaring identity (3.2) finally setting v = mk. Corollary 3.4. If n is an integer, then L 4 k = 5F 2n+1 F n 1 F n+2 +6n 5. Proof. From Theorem 3.3 we have L 4 k = F 2n+1 (L 2n+1 4( 1) n 1 )+6n 5. (3.10) From identity (3.15) with u = n+2 v = n 1 we have L 2n+1 4( 1) n 1 = L 2n+1 ( 1) n 1 L 3 = 5F n 1 F n+2, (3.11) the result follows. 3.2 Alternating sums Theorem 3.5. If m n are integers, then ( 1) k 1 F 4 mk = F { } mnf mn+m ( 1) n 1 n n+m +( 1) n(m 1) 4. 5 Proof. Multiplying through identity (3.3) by ( 1) k 1 summing over k, we have the identity ( 1) k 1 F 4 mk = ( 1) k 1 L 4mk +4 ( 1) k(m 1) k +3(( 1) n 1 +1). (3.12) 6

7 When Lemma 2.4 is used to evaluate the sums on the right h side we have ( 1) k 1 F 4 mk = ( 1)n 1 L 4mn+2m + + 4{ } ( 1) n(m 1) n+m (3.13) +3 { ( 1) n 1 +1 }, that is, ( 1) k 1 F 4 mk = ( 1)n 1 L 4mn+2m = ( 1)n 1 {L 4mn+2m } + 4( 1)n(m 1) n+m + 4( 1)n(m 1) {n+m ( 1) mn }. Theorem 3.5 then follows when the identities +3( 1) n 1 (3.14) L u+v ( 1) v L u v = 5F v F u (3.15) are used to write the right h side of (3.14). F 2u = F u L u (3.16) Corollary 3.6. If n is an integer, then ( 1) k 1 F 4 k = ( 1)n 1 F n F n+1 F n 2 F n+3. 3 Proof. From Theorem 3.5 ( 1) k 1 F 4 k = F nf n+1 (( 1) n 1 L n L n+1 +L 2 L 3 ). (3.17) 15 From identity (3.15) L n L n+1 = L 2n+1 ( 1) n 1, L 2 L 3 = L (3.18) 7

8 We therefore have ( 1) k 1 F 4 k = ( 1)n 1 F n F n+1 (L 2n+1 +( 1) n 1 L 5 ) 15 = ( 1)n 1 F n F n+1 (L 2n+1 ( 1) n 2 L 5 ) 15 = ( 1)n 1 F n F n+1 F n 2 F n+3, by identity (3.15). 3 Theorem 3.7. If m n are integers, then k=(1+( 1) n )/2 ( 1) k 1 k 4 = ( 1)n 1 5F mn F mn+m { n n+m +( 1) nm 4 }. Proof. Multiplying through identity (3.9) by ( 1) k 1 summing over k, we have the identity ( 1) k 1 L 4 mk = ( 1) k 1 L 4mk 4 ( 1) k(m 1) k +3(( 1) n 1 +1), (3.19) which by the use of Lemma 2.4 gives ( 1) k 1 L 4 mk = ( 1)n 1 L 4mn+2m 4( 1)n(m 1) n+m +3( 1) n 1 +8, so that if n is even we have ( 1) k 1 L 4 mk = (L 4mn+2m ) 4(n+m ), (3.20) while if n is odd we have ( 1) k 1 L 4 mk = (L 4mn+2m ) 4( 1)m 1 (n+m ( 1) m ) +16, 8

9 that is, k=0 ( 1) k 1 k 4 = (L 4mn+2m ) 4( 1)m 1 (n+m ( 1) m ). (3.21) Using identities (3.15) (3.16) to write the right side of identities (3.20) (3.21) combining the results we obtain the statement of Theorem 3.7. Corollary 3.8. If n is an integer, then k=(1+( 1) n )/2 References ( 1) k 1 L 4 k = ( 1) n 15 3 F nf n+1 (L n 2 L n+3 +( 1) n 2). [1] S. CLARY P. D. HEMENWAY (1993), On sums of cubes of Fibonacci numbers, in Applications of Fibonacci Numbers, Kluwer Academic Publishers, Dordrecht, The Netherls 5: [2] R. S. MELHAM (2000), Alternating sums of fourth powers of Fibonacci Lucas numbers, The Fibonacci Quarterly 38 (3):4 9. [3] K. ADEGOKE (2017), Generalizations for reciprocal Fibonacci- Lucas sums of Brousseau, arxiv: