aa r X i v : . [ m a t h . N T ] M a y Sums of fourth powers of Fibonacci andLucas numbers ∗ Kunle Adegoke † Department of Physics and Engineering Physics,Obafemi Awolowo University, Ile-Ife, Nigeria
Abstract
We obtain closed-form expressions for all sums of the form P nk =1 F mk and P nk =1 L mk and their alternating versions, where F i and L i de-note Fibonacci and Lucas numbers respectively. Our results com-plement those of Melham who studied the alternating sums. The Fibonacci numbers, F n , and Lucas numbers, L n , are defined, for n ∈ Z , as usual, through the recurrence relations F n = F n − + F n − , F = 0 , F = 1 and L n = L n − + L n − , L = 2 , L = 1 , with F − n = ( − n − F n and L − n = ( − n L n .About two decades ago, motivated by the results of Clary and Hemen-way [1] who obtained factored closed-form expressions for sums of theform P nk =1 F mk , Melham [2] obtained factored closed-form expressionsfor alternating sums of the form P nk =1 ( − k − F mk . ∗ AMS Classification Numbers : 11B37, 11B39 † [email protected], [email protected] m and n , with m not equal tozero: n X k =1 F mk = F mn + m ( L mn + m + 4( − mn − L m ) F m + 6 n + 3 and n X k =1 L mk = F mn + m ( L mn + m + 4( − mn L m ) F m + 6 n − . We also re-derived the alternating sums, in slightly different but equivalentforms to the results contained in [2]: n X k =1 ( − k − F mk = F mn F mn + m (cid:8) ( − n − L m L mn L mn + m + ( − n ( m − L m (cid:9) L m L m and n X k =(1+( − n ) / ( − k − L mk = ( − n − F mn F mn + m { L m L mn L mn + m + ( − nm L m } L m L m , valid for all integers m and n . The following telescoping summation identities are special cases of themore general identities proved in [3].
Lemma 2.1. If f ( k ) is a real sequence and m and n are positive integers,then n X k =1 [ f ( mk + m ) − f ( mk )] = f ( mn + m ) − f ( m ) . Lemma 2.2. If f ( k ) is a real sequence and m and n are positive integers,then n X k =1 ( − k − [ f ( mk + m ) + f ( mk )]= ( − n − f ( mn + m ) + f ( m ) . .2 First-order Lucas summation identities Lemma 2.3. If m and n are integers, then F m n X k =1 ( − mk − L mk = ( − mn − F mn L mn + m . Proof.
Setting v = m and u = 2 mk in the identity F u + v − ( − v F u − v = F v L u , (2.1)gives F mk + m − F mk − m = F m L mk , m even , (2.2)and F mk + m + F mk − m = F m L mk , m odd . (2.3)Using identity (2.2) in Lemma 2.1 with f ( k ) = F k − m , it is establishedthat F m n X k =1 L mk = F m +2 mn − F m = F m + mn + mn − F m + mn − mn = F mn L mn + m , m even , (2.4)on account of identity (2.1).Similarly, using identity (2.3) in Lemma 2.2 with f ( k ) = F k − m , we have F m n X k =1 ( − k − L mk = ( − n − F m +2 mn + F m = ( − n − ( F m + mn + mn − ( − n F m + mn − mn )= ( − n − ( F m + mn + mn − ( − mn F m + mn − mn ) , since m is odd = ( − n − F mn L mn + m , m odd . (2.5)Identities (2.4) and (2.5) combine to give Lemma (2.3). Lemma 2.4. If m and n are integers, then L m n X k =1 ( − k ( m − L mk = ( − n ( m − L mn + m − L m . roof. Setting v = m and u = 2 mk in the identity L u + v + ( − v L u − v = L v L u , (2.6)gives L mk + m − L mk − m = L m L mk , m odd , (2.7)and L mk + m + L mk − m = L m L mk , m even . (2.8)Using (2.7) in Lemma 2.1 with f ( k ) = L k − m , we have L m n X k =1 L mk = L m +2 mn − L m , m odd . (2.9)Similarly, using (2.8) in Lemma 2.2 with f ( k ) = L k − m , we have L m n X k =1 ( − k − L mk = ( − n − L m +2 mn + L m m even . (2.10)Identities (2.9) and (2.10) combine to give Lemma (2.4). Theorem 3.1. If m is a non-zero integer and n is any integer, then n X k =1 F mk = F mn + m ( L mn + m + 4( − mn − L m ) F m + 6 n + 3 . Proof.
By squaring the identity F u = L u − ( − u , u ∈ Z , (3.1)and making use of the identity L v = L v + ( − v , v ∈ Z , (3.2)and finally setting u = mk , it is established that F mk = L mk + ( − mk − L mk + 6 . (3.3)4y summing both sides of identity (3.3), using Lemma 2.3 to sum each ofthe first two terms on the right hand side, we have n X k =1 F mk = ( F mn L mn +2 m + 4( − mn − L m F mn L mn + m ) F m + 6 n , (3.4)Using the identity (2.1) we can write F mn L mn +2 m = F mn +2 m − F m = F mn + m L mn + m − F m (3.5)and L m F mn L mn + m = L m ( F mn + m − ( − mn F m )= L m F mn + m + ( − mn − F m . (3.6)Substituting (3.5) and (3.6) into (3.4) proves Theorem 3.1. Corollary 3.2. If n is an integer, then n X k =1 F k = F n +1 L n − L n +2 + 6 n + 3 . Proof.
From Theorem 3.1 we have n X k =1 F k = F n +1 ( L n +1 + 4( − n − ) + 6 n + 3 . (3.7)From identity (2.6) with u = n + 2 and v = n − we have L n +1 + 4( − n − = L n +1 + ( − n − L = L n − L n +2 , (3.8)and the result follows. Theorem 3.3. If m is a non-zero integer and n is any integer, then n X k =1 L mk = F mn + m ( L mn + m + 4( − mn L m ) F m + 6 n − . roof. The theorem is proved by summing both sides of the followingidentity, L mk = L mk − ( − mk − L mk + 6 , (3.9)applying Lemma 2.3 to sum each of the first two terms on the right handside. Identity (3.9) is obtained by squaring identity (3.2) and finally setting v = mk . Corollary 3.4. If n is an integer, then n X k =1 L k = 5 F n +1 F n − F n +2 + 6 n − . Proof.
From Theorem 3.3 we have n X k =1 L k = F n +1 ( L n +1 − − n − ) + 6 n − . (3.10)From identity (3.15) with u = n + 2 and v = n − we have L n +1 − − n − = L n +1 − ( − n − L = 5 F n − F n +2 , (3.11)and the result follows. Theorem 3.5. If m and n are integers, then n X k =1 ( − k − F mk = F mn F mn + m (cid:8) ( − n − L m L mn L mn + m + ( − n ( m − L m (cid:9) L m L m . Proof.
Multiplying through identity (3.3) by ( − k − and summing over k , we have the identity n X k =1 ( − k − F mk = n X k =1 ( − k − L mk + 4 n X k =1 ( − k ( m − L mk + 3(( − n − + 1) . (3.12)6hen Lemma 2.4 is used to evaluate the sums on the right hand side wehave n X k =1 ( − k − F mk = ( − n − L mn +2 m + L m L m + 4 (cid:8) ( − n ( m − L mn + m − L m (cid:9) L m + 3 (cid:8) ( − n − + 1 (cid:9) , (3.13)that is, n X k =1 ( − k − F mk = ( − n − L mn +2 m L m + 4( − n ( m − L mn + m L m + 3( − n − = ( − n − { L mn +2 m − L m } L m + 4( − n ( m − { L mn + m − ( − mn L m } L m . (3.14)Theorem 3.5 then follows when the identities L u + v − ( − v L u − v = 5 F v F u (3.15)and F u = F u L u (3.16)are used to write the right hand side of (3.14). Corollary 3.6. If n is an integer, then n X k =1 ( − k − F k = ( − n − F n F n +1 F n − F n +3 . Proof.
From Theorem 3.5 n X k =1 ( − k − F k = F n F n +1 (( − n − L n L n +1 + L L )15 . (3.17)From identity (3.15) L n L n +1 = L n +1 − ( − n − , L L = L + 1 . (3.18)7e therefore have n X k =1 ( − k − F k = ( − n − F n F n +1 ( L n +1 + ( − n − L )15= ( − n − F n F n +1 ( L n +1 − ( − n − L )15= ( − n − F n F n +1 F n − F n +3 , by identity (3.15). Theorem 3.7. If m and n are integers, then n X k =(1+( − n ) / ( − k − L mk = ( − n − F mn F mn + m { L m L mn L mn + m + ( − nm L m } L m L m . Proof.
Multiplying through identity (3.9) by ( − k − and summing over k , we have the identity n X k =1 ( − k − L mk = n X k =1 ( − k − L mk − n X k =1 ( − k ( m − L mk + 3(( − n − + 1) , (3.19)which by the use of Lemma 2.4 gives n X k =1 ( − k − L mk = ( − n − L mn +2 m L m − − n ( m − L mn + m L m +3( − n − +8 , so that if n is even we have n X k =1 ( − k − L mk = − ( L mn +2 m − L m ) L m − L mn + m − L m ) L m , (3.20)while if n is odd we have n X k =1 ( − k − L mk = − ( L mn +2 m − L m ) L m − − m − ( L mn + m − ( − m L m ) L m + 16 , n X k =0 ( − k − L mk = − ( L mn +2 m − L m ) L m − − m − ( L mn + m − ( − m L m ) L m . (3.21)Using identities (3.15) and (3.16) to write the right side of identities (3.20)and (3.21) and combining the results we obtain the statement of Theo-rem 3.7. Corollary 3.8. If n is an integer, then n X k =(1+( − n ) / ( − k − L k = ( − n − F n F n +1 ( L n − L n +3 + ( − n . References [1] S. CLARY and P. D. HEMENWAY (1993), On sums of cubes ofFibonacci numbers, in Applications of Fibonacci Numbers, KluwerAcademic Publishers, Dordrecht, The Netherlands
The Fibonacci Quarterly
38 (3):254–259.[3] K. ADEGOKE (2017), Generalizations for reciprocal Fibonacci-Lucas sums of Brousseau, arXiv:1702.08321 https://arxiv.org/abs/1702.08321https://arxiv.org/abs/1702.08321