An Identity in the Generalized Fibonacci Numbers and Its Applications

We first generalize an identity involving the generalized Fibonacci numbers and then apply it to establish some general identities concerning special sums. We also give a sufficient condition on a generalized Fibonacci sequence {Un} such that Un is divisible by an arbitrary prime r for some 2 < n ≤ r − 2. 1. Preliminaries The generalized Fibonacci and Lucas numbers are defined, respectively, by Binet’s formula, as follows Un(p, q) = αn − βn α− β , Vn(p, q) = α n + β, where α = 2 (p + √ p2 − 4q) and β = 2 (p− √ p2 − 4q). The numbers Un(p, q) and Vn(p, q) can be defined recursively by Un(p, q) = −qUn−2(p, q) + pUn−1(p, q), Vn(p, q) = −qVn−2(p, q) + pVn−1(p, q), for all integers n, where U0 = 0, U1 = 1, V0 = 2 and V1 = p. Throughout the paper, p and q denote the real numbers, Un and Vn stand for Un(p, q) and Vn(p, q), respectively, and ∆ = p2 − 4q. A sequence {Gn} is said to be a (p, q)-sequence if Gn satisfies the recursive relation Gn = −qGn−2 + pGn−1, for all integers n. Clearly, the (p, q)-sequences, which are identified at two consecutive indices should be equal. It is known that the formula Ua+b = −qUa−1Ub + UaUb+1 INTEGERS: 9 (2009) 498 is valid for any generalized Fibonacci sequence {Un(p, q)} and all integers a, b. We intend to present a generalization of this identity and derive some of its applications, which are the general solutions of some solved and unsolved problems concerning the generalized Fibonacci numbers. In Section 2, we prove our claim and give its generalization. In Section 3, we apply our identity to evaluate some summations involving that of Mansour [4] (see also [5]), the sum of powers of the generalized Fibonacci numbers and etc. In final section, we use our identity to get a divisibility property of the generalized Fibonacci numbers. Remark 1. In the sequel we shall frequently use the fact that if a finite rational expression P contains some terms of a generalized Fibonacci sequence, which is not identically zero but it vanishes with respect to a special sequence {Un}, we can always choose a sequence of the generalized Fibonacci sequences {Um n }m=1 such that P does not vanishes over these sequences, while {Um n }m=1 tends to {Un}. Without loss of generality, we may assume that all the sequences under the consideration do not vanish over the expressions, which might appear in the denominators. 2. Main Results It is well-known that if {Un(p, q)} is a generalized Fibonacci sequence, then Ua+b = −qUa−1Ub + UaUb+1, for all integers a, b. It is also proved in [2, Lemma 2.1(c)] that the identity Fa+b+c−3 = FaFbFc + Fa−1Fb−1Fc−1 − Fa−2Fb−2Fc−2 is valid, for all integers a, b, c. We give a generalization of these identities in terms of the generalized Fibonacci sequences. Theorem 2. If {Un} is a generalized Fibonacci sequence, then for all natural numbers m, Ua1+···+am−(m+1 2 ) = m ∑ i=1 { m i } Ua1−i · · ·Uam−i, (1) where a1, . . . , am are integers and { m i } =   m ∏ j=1 j !=i Uj−i   −1 . Proof. First suppose that a1, . . . , am are equal to 1, 2, . . . , i, i + 2, i + 3, . . . ,m + 1, in some order. Without loss of generality, we may assume that a1 = 1, . . . , ai = i, INTEGERS: 9 (2009) 499 ai+1 = i+2, . . . , am = m + 1. If j #= i+1, then Ua1−j · · ·Uam−j = 0 holds and if j = i+1, then Ua1−j · · ·Uam−j = U−i · · ·U−1U1 · · ·Um−i so that {m j } Ua1−j · · ·Uam−j = Um−i. On the other hand, Ua1+···+am−(m+1 2 ) = Um−i and in this case the equality holds. If Un = Un(p, q), then clearly the both sides of (1) are (p, q)-sequences with respect to each ai and also they identify on the cube (1, 2, . . . ,m) + {0, 1}m. Hence the both sides of (1) should be equal over all the integer values of a1, . . . , am. The proof is now complete. ! Theorem 2 can be generalized in the following manner. Theorem 3. If {Un} is a generalized Fibonacci sequence, then for each natural numbers m and n (with the same parity), Ua1+···+an−(m+1 2 ) = 1 ∆ 2 (m−n) m ∑ i=1 { m i } Ua1−m1i · · ·Uan−mni, where m = m1 + · · · + mn and m1, . . . ,mn are odd natural numbers. Proof. Let Uk = Uk(p, q) and m,m1, . . . ,mn be natural numbers such that m = m1 + · · ·+ mn and m1, . . . ,mn are odd. By putting am−mn+1 = · · · = am = k in Theorem 2, we get Ua1+···+am−mn+mnk−(m+1 2 ) = m ∑ i=1 { m i } Ua1−i · · ·Uam−mn−iU mn k−i. (2) By definition, Uk = (αk−βk)/(α−β), where α = (p+ √ ∆)/2 and β = (p− √ ∆)/2. Hence if ∆ > 0, then β < α and so limk→∞ Uk/α = 1/(α−β), from which together with (2) we obtain (α− β)mn−1αa1+···+am−mn−( m+1 2 ) = m ∑ i=1 { m i } Ua1−i · · ·Uam−mn−iα −mni. (3) A simple computation shows that αk = −qUk−1 +Ukα, for each integer k. Suppose that p and q are rational but α is irrational. This together with (3) yields (α− β)Ua1+···+am−mn+(m+1 2 )−1 = m ∑ i=1 { m 2 } Ua1−i · · ·Uam−mn−iU−mni−1