ADVANCED PROBLEMS AND SOLUTIONS

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1 ADVANCED PROBLEMS AND SOLUTIONS EDITED BY FLORIAN LUCA Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS to FLORIAN LUCA, MATHEMATICAL INSTITUTE, UNAM, CP 0450, MEXICO DF, MEXICO or by at as files of the type tex, dvi, ps, doc, html, pdf, etc. This department especially welcomes problems believed to be new or extending old results. Proposers should submit solutions or other information that will assist the editor. To facilitate their consideration, all solutions sent by regular mail should be submitted on separate signed sheets within two months after publication of the problems. PROBLEMS PROPOSED IN THIS ISSUE H-75 Proposed by D. M. Bătineţu-Giurgiu, Bucharest and Neculai Stanciu, Buzău, Romania. Prove that F 2n m+ + 2n+ F 2n m+ F m + + 2n+ n F n m+ Fm n + 2n+ n+ F m F n m+ F n m + + 2n+ 2n F m+ Fm+ F 2n m p+ p+ F 2n+ p+ m+2 2 p F m F m+ holds for any p 0 and positive integers m and n, and that the same inequality holds with all the F s replaced by L s. H-752 Proposed by D. M. Bătineţu-Giurgiu, Bucharest and Neculai Stanciu, Buzău, Romania. Prove that 2n+ 2n+ p 2m+ p 2m+ p p 5 m L 2m+ F 5 n L 2n+ F, p p p0 2n+ 2 5 m F 2m+ p0 0 p 2n+ p 0 p p0 0 2m+ p 2m+ L 5 n F 2n+ p p0 0 p L. H-753 Proposed by H. Ohtsua, Saitama, Japan. For integers n, m, a 0 and b, prove that n 2m Fa+b 4m 4m an+b+r F anr L an+a+2br 2mr 25 m + F ar r 4m n 2m 25 m. MAY

2 THE FIBONACCI QUARTERLY H-754 Proposed by H. Ohtsua, Saitama, Japan. Let a, b and n be integers. The two sequences {T n } and {S n } satisfy T n+3 T n+2 +T n+ +T n with arbitrary T 0, T, T 2, S n+3 S n+2 +S n+ +S n with S 0 3, S, S 2 3 for all integers n. Let R n S n +. For n, prove that n Ra 2 R2 a Ta+b 2 A n A 0, where A n 2T an+b R a T an+a+b +R a T ana+b T an+a+b T ana+b 2 R a T an+b 2. SOLUTIONS Infinite Sums With Reciprocals of Squares of Fibonacci and Lucas Numbers H-724 Proposed by H. Ohtsua, Saitama, Japan. Vol. 50, No. 3, August 202 Determine Solution by the proposer. F 2 4 The following identity is easily verified. L L 2 2 5F 2 n L2 n 4n. If n 0, dividing both sides of the identity by F 2 2n, 2 5 L 2 n F 2 n 4n F2n 2. F 2 2. Using the above identity, we have F 2 L 2 + L 2 F F L 2 + L 2 F F 2 F F F F 2 + F 2 4 F F 2 4 F VOLUME 52, NUMBER 2

3 ADVANCED PROBLEMS AND SOLUTIONS Thus, we have Therefore, 5 F 2 4 F 2 4 L L L 2 2 L 2 2 F 2 2 F Also solved by Paul S. Brucman and Dmitry Fleishman. Sums With Powers of 3, 4 and Binomial Coefficients H-725 Proposed by Paul S. Brucman, Nanaimo, BC. Vol. 50, No. 4, November 202 Prove the following identities valid for n 0,,2,... n3 a n4 6 0 where sinθ 2/3; n b n4 8 3n+53 n n 3 n/2sinnθ sinθ, 9n+7+3 3n/2 cosnρ+sinnρ/ 2, where sinρ 2/27; n2 pqp 2 q 2 2 c 2 p 2 +q 2 2 pp+qn+ qq p n+ 2p 2 +2pq q 2 p 2 +q 2 n 0 + ppqn+ qq +p n+ 2p 2 2pq q 2 p 2 +q 2, where p > q > 0 are integers. n Solution by G. C. Greubel. a The process for the three series will be to consider the generating function of the given relations and compare the final results. For the first series in question consider From this, we have Sn tn n,0 0 S n n+ 0 n n t n n t t 4 4t 4t 27t 4 0 4t + n 4t n n! 4t+27t4. MAY

4 THE FIBONACCI QUARTERLY Alternatively, let φ n be given by for which φ n 6 [ φ n tn 6 3n+53 n n 3 n/2sinnθ, sinθ 56t 3t 2 ] 3t n sinnθ. sinθ The summation on the right-hand side can be evaluated as follows. 3t n sinnθ e inθ e inθ 3t n 2i e iθ 2i + 3e e iθ iθt cosθt sinθ 3e iθ t +2. 3cosθt+3t 2 From this, we have [ φ nt n ] 56t 6 3t cosθt +2. 3cosθt+3t 2 In order to reduce this expression use sinθ 2/3 or cosθ /3 to obtain φ n tn [ ] 56t 6 6t+9t t +2t+3t 2 By comparing equations and 2, we are led to the desired result 0 n3 where sinθ 2/ n4 6 4t+27t2. 2 3n+53 n n 3 n/2sinnθ, sinθ b Let Sn 2 be the series to be summed. As before, consider the generating function for this series. Snt 2 n [n/4] 0 n t n4 t n n,0 3 + n 4t n n! 4t n n t n+3 27t 4 0 4t 3 4t 2 4t 3 +27t VOLUME 52, NUMBER 2

5 ADVANCED PROBLEMS AND SOLUTIONS Now that we have the desired generating function to compare to, let φ 2 n be the right-hand side of the desired result and proceed to find its generating function. φ 2 n tn 7+9n+ 33n/2 2cosnρ+sinnρ t n 8 2 where Υ,2 are the series Υ 8 7+2t t 2 + Υ + Υ 2, cosnρ3 3/2 t n, and Υ 2 Evaluating Υ yields Υ [ ] 2 3 3/2 e iρ t + 3 3/2 e iρ t The result for Υ 2 is similar and is given by Υ 2 sinnρ3 3/2 t n. 3 3/2 cosρ t 2 3 3/2 cosρ t+27t /2 sinρ t 2 3 3/2 cosρ t+27t2. 6 By combining 4, 5 and 6, the resulting series taes the form φ 2 n tn 7+2t 8 t /2 2cosρsinρt /2 cosρt+27t 2 Now using the provided value sinρ 2/27, which also provides cosρ 25/27, leads to the reduction φ 2 nt n [ ] 7+2t 8 t t 0t+27t 3 [ 844t+268t 2 ] 8 2t+48t 2 64t 3 +27t 4 4t 2 4t 3 +27t4. 8 Since 3 and 8 have the same generating function, then we conclude that 0 n 3 where sinρ 2/ n n+ 33n/2 2cosnρ+sinnρ, 9 2 c Let x pqp2 q 2 p 2 +q 2 2, 0 MAY

6 THE FIBONACCI QUARTERLY and let Sn 3 be the series in question, then [n/4] n2 Sn 3 tn 2 0 n,0 x 2 t n n,0 2 + n x 2 t 4 t n n! t n+2 2 x 2 t 4 0 x 2 t 4 t n t 2 t t 2 x 2 t4. Alternatively, let φ 3 n be the right-hand side of the third series in question and see the generating function. where Υ 3 φ 3 nt n 2p 2 +2pq q 2 Υ 3 + pp+q n+ qq p n+ p 2 +q 2 n t n, and Υ 4 2p 2 2pq q 2 Υ 4, 2 ppq n+ qp+q n+ p 2 +q 2 n t n. Consider the evaluation of Υ 3. This is done in the following way. pp+q n+ qq p n+ Υ 3 p 2 +q 2 n t n pp+qp2 +q 2 p 2 +q 2 pp+qt qq pp2 +q 2 p 2 +q 2 qq pt p 2 +2pq q 2 p 2 +q 2 2 p 2 +q 2 2 tpqp 2 q 2 t 2 p2 +2pq q 2 txt 2, 3 where x is given by 0. The same process may be applied to evaluating Υ 4 and we are led to the result Υ 4 p2 2pq q 2 t+xt 2. 4 From these expressions equation 2 is reduced to φ 3 n tn 2 txt 2 + t+xt 2 t t 2 x 2 t 4. This is the same generating function as that of equation. Thus leading to the statement n2 pqp 2 q p 2 +q 2 2 pp+qn+ qq p n+ 2p 2 +2pq q 2 p 2 +q 2 n 0 where p > q > 0 are integers. + ppqn+ qq +p n+ 2p 2 2pq q 2 p 2 +q 2 n, 5 Also solved by Paul S. Brucman and Kenneth B. Davenport. 90 VOLUME 52, NUMBER 2

7 ADVANCED PROBLEMS AND SOLUTIONS Sums of Sums of Reciprocals of Fibonacci Numbers H-726 Proposed by Hideyui Ohtsua, Saitama, Japan. Vol. 50, No. 4, November 202 Prove that F 2 F 4 F 8 F 6 F 2 n F 2 F 2. Solution by the proposer. First, we will prove the following lemma. Lemma. F n F n+ α n + F n α n+ ; F n+ 2 F 2n α n n F n α 2n. F 2n Proof of Lemma. We have Thus, 5αFn+ +F n αα n+ β n+ +α n β n α n+2 +α n α n α 2 + 5α n+. α n+ αf n+ +F n. Dividing both sides of this identity by α n+ F n F n+, we get identity. 2 We have Thus, α n L n α n α n +β n α 2n + n. α 2n α n L n n. Dividing both sides of this identity by α 2n F 2n, we get identity 2. Using Lemma, we have F 2 F 2 Using Lemma 2, we have F 4 α 2 + F 2 α 2 F 2 α 2 F 2 α 4 F 4 α F. α 42 F MAY 204 9

8 THE FIBONACCI QUARTERLY Using Lemma 2, we have We have Thus, we obtain, F n3 2 n lim N F 2 lim N n3 α 4 F 4 F 2 F 2 F 2 F 2 F 2 α F N α 2n F 2 n α 2n F 2 n α 2N F 2 N α 4 F 4 + F n3 2 n F 4 + Also solved by Paul S. Brucman. α 4 F 4. 3 α 2 F 2 by Lemma 2 + F n3 2 n α 42 F 42 by F 4. by 2 and 3. F 2 F VOLUME 52, NUMBER 2