ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS EDITED BY HARRIS KWONG Please submit solutions and problem proposals to Dr Harris Kwong, Department of Mathematical Sciences, SUNY Fredonia, Fredonia, NY 4063, or by at If you wish to have receipt of your submission acknowledged by mail, please include a selfaddressed, stamped envelope Each problem or solution should be typed on separate sheets Solutions to problems in this issue must be received by August 5, 07 If a problem is not original, the proposer should inform the Problem Editor of the history of the problem A problem should not be submitted elsewhere while it is under consideration for publication in this Journal Solvers are asked to include references rather than quoting well-known results The content of the problem sections of The Fibonacci Quarterly are all available on the web free of charge at wwwfqmathca/ Dedication: The problems in this issue are dedicated to Dr Russ Euler and Dr Jawad Sadek in recognition of their sixteen years of devoted service as the editors of the Elementary Problems and Solutions Section BASIC FORMULAS The Fibonacci numbers F n and the Lucas numbers L n satisfy F n = F n F n, F 0 = 0, F = ; L n = L n L n, L 0 =, L = Also, α = 5/, β = 5/, F n = α n β n / 5, and L n = α n β n PROBLEMS PROPOSED IN THIS ISSUE B-0 Proposed by Ivan V Fedak, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine If a,b,c > 0, then prove that, for any positive integer n, a 3 af n bf n cf n a 3 al n bl n cl n a 3 b 3 a b, af n bf n bf n af n F n a 3 b 3 a b, al n bl n bl n al n L n b 3 bf n cf n af n b 3 bl n cl n al n c 3 a b c, cf n af n bf n F n c 3 a b c cl n al n bl n L n 8 VOLUME 55, NUMBER

2 ELEMENTARY PROBLEMS AND SOLUTIONS B-0 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania, Neculai Stanciu, George Emil Palade School, Buzău, Romaina, and Gabriel Tica, Mihai Viteazul National College, Băileşti, Dolj, Romania a n Let a n n be a positive real sequence such that lim n n = a Evaluate a n a n n F n lim n an F n a n n L n and lim n an L n n n! n! n n! n! B-03 Proposed by Hideyuki Ohtsuka, Saitama, Japan Prove that, for any positive integer n, i F F3k F F3k F F3k = 3 F Fk ; 5 ii L L3k L L3k L L3k = n 3 L k L Lk B-04 Proposed by Steve Edwards, Kennesaw State University, Marietta, GA For non-negative integers n, express nj nj i nj A n = j n j i j=0 i=0 in terms of Fibonacci numbers and B n = n j=0 j nj nj i nj n j i i=0 B-05 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania, and Neculai Stanciu, George Emil Palade School, Buzău, Romania Prove that for any positive integers n and m n m F m k F m n Fm n SOLUTIONS Adding Two Terms at a Time B-8 Proposed by Hideyuki Ohtsuka, Saitama, Japan Vol 54, February 06 Prove that n= F n F n = 5 3 FEBRUARY 07 83

3 THE FIBONACCI QUARTERLY Solution by Jeremiah Bartz, University of North Dakota, Grand Forks, ND We make use of the following four identities: 3F i = Fi F i Fi = F i F i3 3 F i = Fi F i See [, p 79] 4 For even i, Fi = F i F i Special case of Cassini s Formula [, p 74] The first identity can be derived as follows From repeated application of the relation F i = F i F i, it follows that F i = F i F i F i, or equivalently, 3F i = F i F i Then using the identities F i = F i L i and L i = F i F i see [, p 80], we see that 3F i = 3F i L i = F i F i F i F i = Fi Fi Identity is a special case of a result due to Sharpe see Identity 54 of [, p 90] Using 4, the given series can be expressed as a telescoping series For M, we have n= Therefore, F n Fn = F F F 4 lim F n= n = lim = lim = lim F M = lim F n N N n=3 F n F n 3Fk 3Fk F k Fk M F k3 Fk M 3F 3F = 3F k F k3 F k3 F k F k F k3 3F k 3F 3 3F 3 3F k3 F 3 F 3 References Fk Fk3 3F M 3F M3 = 3 4 = 5 3 [] T Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, NY, 00 FM3 Also solved by Brian Bradie, Steve Edwards, Harris Kwong, Poo-Sung Park, Jason L Smith, and the proposer Vieta s Formula B-8 Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Vol 54, February 06 Let n be a positive integer If a, b, c are the roots of the equation x 3 F n x F n = 0, then prove that a 3 F n ab 3 F n bc 3 F n c is a positive integer which is the sum of n squares 84 VOLUME 55, NUMBER

4 Solution by Hideyuki Ohtsuka, Saitama, Japan For any root r of the equation x 3 F n x F n = 0 we have ELEMENTARY PROBLEMS AND SOLUTIONS r 3 F n r = rr F n r 3 = rf n In addition, Vieta s formula asserts that abc = F n Using the above identity, we find a 3 F n ab 3 F n bc 3 F n c = abcf n = F n F n = Fk Thus, the given sum is a positive integer which is the sum of n squares Editor s Remark: A similar result holds when the Fibonacci numbers in the problem are replace by the Lucas numbers The sum equals to L n L n = n L k Also solved by Adnan Ali student, Brian D Beasley, Charles K Cook, Itzal De Urioste student, Steve Edwards, I V Fedak, Dmitry Fleischman, G C Gruebel, Russell Jay Hendel, Harris Kwong, Ángel Plaza, Vincelot Ravoson, Jason L Smith, David Terr, and the proposer A Binomial Series B-83 Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain Vol 54, February 06 Express as a function of F k and L k n k L n 5F n Solution by Adnan Ali student, AECS-4, Mumbai, India First, we note that L n 5F n = α n, and so the sum to be evaluated becomes n n k L n = α n 5F n k We know that and so n x n = k x k x k, n α α n k = k α = α k = L k 5F k Also solved by Brian Bradie, Steve Edwards, G C Greubel, Harris Kwong, Hideyuki Ohtsuka, S S Pradhan and M K Sahukar, Jason L Smith, David Terr, and the proposer FEBRUARY 07 85

5 THE FIBONACCI QUARTERLY Matrices of Ones B-84 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzău, Romaina Vol 54, February 06 Let k be a positive integer Fk F If Ak = k, compute F k F k F If Bk = k Fk, compute 3 If Ck = F k F k Ak Bk Fk F k Lk L k, compute F k F k L k L k Solution by Vincelot Ravoson, Paris, France Let J = a b Notice that J b a Let us prove by induction that Ak = Bk = abj for any real numbers a and b n F k J F F For n =, we have A = = J = F F F F F 3 J Now, we assume that the result is true for a fixed positive integer n Then n n n Ak = Ak An = F k JAn Using the remark above, Thus, JAn = n and the conclusion follows Ak = Let us prove by induction that Fn F n = F F n F n F n J = F n3 J n n F k Bk = F n3 J = F k J n3 F k J, 86 VOLUME 55, NUMBER

6 ELEMENTARY PROBLEMS AND SOLUTIONS The proof is essentially the same as the proof of, except that, in the inductive step, we use the identity Fn F n = F n3 to prove that F JBn = n F n Fn Fn = Fn FnJ = F n3 J 3 Let us prove by induction that Ck = n 6 F k J The proof is almost identical to those of and, except that, in the inductive step, Fn F JCn = n Ln L n F n F n L n L n Ln L = F n F n J n L n L n = F n F n L n L n J = F n3 L n3 J = F n6 J Also solved by Jeremiah Bartz, Brian Bradie, Steve Edwards, Dmitry Fleischman, Hideyuki Ohtsuka, Ángel Plaza, Jason L Smith, David Stone and John Hawkins, David Terr, and the proposer Chebyshev, Jensen, or Nesbitt? B-85 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzău, Romania Vol 54, February 06 Let a and b be positive real numbers such that af n > bf n Prove that Fk af n F n bfk n an b for all n Solution by Hideyuki Ohtsuka, Saitama, Japan Note that for k n, af n F n bf k af nf n bf n = F n af n bf n > 0, and for k n, we have Fk F k and af n F n bfk af n F n bfk By Chebyshev s inequality and the AM-HM inequality, Fk af n F n bfk Fk n n n af n F n bfk F k n af nf n bfk FEBRUARY 07 87

7 THE FIBONACCI QUARTERLY Note that n F k = F nf n, we have F k af n F n bf k nf n F n = n anf n F n bf n F n an b Solution by Brian Bradie, Christopher Newport University, Newport News, VA Given af n > bf n, it follows that af n F n > bfn bf k Let fx = x/af nf n bx Then f abf n F n x = af n F n bx > 0 for x [ F n],f, so f is convex in this interval Now, Jensen s inequality yields Fk af n F n bfk = ffk n f Fk n Since n F k = F nf n, F k af n F n bf k n Solution 3 by G C Greubel, Newport News, VA n F nf n af n F n b n F = n nf n an b By making use of the generalized Nesbitt s inequality see, for example, [] x α α k a c n n j= x n α x k j bxk anc b for the case of x k = F k and c = 0, the given inequality becomes F α k af n F n bfk n α F n F n α an b When α = 0 the desired result is obtained, namely, When α = the inequality becomes F k af n F n bf k F 4 k af n F n bf k References n an b F nf n an b [] M Bencze and O T Pop, Generalizations and refinements for Nesbitt s inequality, J Math Inequal, 5 0, 3 0 Also solved by Adnan Ali student, Ángel Plaza, and the proposer 88 VOLUME 55, NUMBER