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 November 5, 08 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/ 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-6 Proposed by Ángel Plaza and Sergio Falcón, Universidad de Las Palmas de Gran Canaria, Spain For any positive integer k, the k-fibonacci and k-lucas sequences, denoted {n } n 0 and {L k,n } n 0 respectively, are both defined recursively by u n+ ku n + u n for n, with initial conditions,0 0,,, and L k,0, L k, k Prove that, for any positive integer n, Fk,j,j,n,n+ k B-7 Proposed by Kenny B Davenport, Dallas, PA j Let C n n+( n n denote the nth Catalan number Find the closed form expressions for the sums n0 C n F n 8 n and n0 C n L n 8 n MAY 08 77

2 THE FIBONACCI QUARTERLY B-8 Proposed by Hideyuki Ohtsuka, Saitama, Japan For any integer n 0, find the closed form expressions for the sums (i S n L 3 il 3 jl (3 i 3 j ; (ii T n i0 j0 i0 j0 F 5 if 5 jl 3(5 i 5 j B-9 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania, and Neculai Stanciu, George Emil Palade School, Bazău, Romania Let m, p 0 Evaluate n+ ((n m+f p(m+ n ((n m+f p(m+ +!! n+!! n lim n (n + m n m, and lim n n+ ((n m+l p(m+ +!! n+ (n + m n ((n m+l p(m+!! n n m B-30 Proposed by T Goy, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine For all integers n 0, prove that F n+ ( n t,t,,tn 0 t +t + +ntnn ( t +t 3 + +t n [+( n ]/ (t + t + + t n! t! t! t n! t SOLUTIONS Two Doses of AM-GM Inequality B-06 Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain (Vol 55, May 07 Let n be an integer Prove that + ( Fi F j+ F i+ F j n F i F j in which the subscripts are taken modulo n n, Solution by Ivan V Fedak, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine To begin, we rewrite the given inequality in the form 78 VOLUME 56, NUMBER

3 n + Next, we see that ELEMENTARY PROBLEMS AND SOLUTIONS ( F i+ + F j+ F i+ Fj+ n F i F j F j F j ( Fi+ + F j+ F i F j F i+ Fj+ n(n F i F j n(n (n n n n n,, F i+ Fj+ F i F j in which the subscripts are taken modulo n From it, follows that n + ( F i+ + F j+ F i+ Fj+ n + (n F j F i F i F j n + (n n n(n, n(n + (n Also solved by Brian Bradie, Dmitry Fleischman, Hideyuki Ohtsuka, Albert Stadler, and the proposer Bergström on a Cyclic Sum B-07 Proposed by D M Bătineţu-Giurgiu, Matei Basarab National College, Bucharest, Romania, and Neculai Stanciu, George Emil Palade School, Buzău, Romania (Vol 55, May 07 Prove that for any integer n > F 4 n + F 4 F n + F n + F 4 k + F 4 k+ > F n F n+ Solution by Brian Bradie, Christopher Newport University, Newport News, VA Let x and y be positive real numbers Then x + y x + y (x + y (x + y (x + y (x + y (x y (x + y 0, MAY 08 79

4 THE FIBONACCI QUARTERLY so that x + y x + y (x + y, with equality holding if and only if x y Using this inequality and the identity Fk +, it follows that F 4 n + F 4 F n + F n + F 4 k + F 4 k+ F 4 n + F 4 F n + F n + (F n + F + F k F n F n+ F 4 k + F 4 k+ F k + F k+ n ( F k + Fk+ Note that equality holds for n as F F, but the inequality is strict for n > Solution by Wai-Kai Lai and John Risher (student (jointly, University of South Carolina Salkehatchie, Walterboro, SC Since Fk +, and F nf n+ n, the claimed inequality is equivalent to F 4 n + F 4 F n + F + F 4 + F 4 F + F Applying Bergström s inequality, we have F 4 n F n + F F 4 F n + F + F 4 F + F + F 4 F + F + + F 4 n F n + F n + + F 4 n + F 4 n F n + F n F 4 n + + Fn + F n Fk (F + + F n (F + + F n (F + + F n (F + + F n Fk, Fk Summing these two inequalities we then have the desired claim The equality holds only when n Also solved by Kenny B Davenport, I V Fedak, Dmitry Fleischman, Donghae Lee (middle school student, Hideyuki Ohtsuka, Ángel Plaza, Henry Ricardo, Zachery R Smith (student, and the proposer Row Reduction on an Augmented Matrix B-08 Proposed by Ivan V Fedak, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine (Vol 55, May VOLUME 56, NUMBER

5 ELEMENTARY PROBLEMS AND SOLUTIONS For every positive integer n, find all real solutions of the following linear system of equations: F x + x F 3, F x + F x + x 3 F 4, F 3 x + F x + F x 3 + F 5, F n x + F n x + F n 3 x x n F n+, F n x + F n x + F n x F x n + x n+ F n+, F n+ x + F n x + F n x F x n + F x n+ F n+3 Composite solution by the proposer and the Elementary Problems Editor We can use Gauss-Jordan elimination to solve the linear system We want to apply row reduction to the following augmented matrix: F F 3 F F 0 0 F 4 F 3 F F 0 0 F 5 F n F n F n 3 0 F n+ F n F n F n F F n+ F n+ F n F n F F F n+3 For k 3, 4,, n +, subtracting the sum of the first k rows from row k reduces the augmented matrix to (recall that F + F + + F m F m For k n, n,,,, subtracting row k from row k + further reduces the augmented matrix to We deduce that x x, x 3 x n, and x k x k+ for k, 3,, n Therefore, if n is even, the solution is x x x n+ ; but if n is odd, the solution is x t, x x 4 x n+ t, x 3 x 5 x n, where t is an arbitrary real number MAY 08 8

6 THE FIBONACCI QUARTERLY Also solved by Hsin-Yun Ching (student, Dmitry Fleischman, and Jason L Smith A Bijective Argument on a Tribonacci Sum B-09 Proposed by Hideyuki Ohtsuka, Saitama, Japan (Vol 55, May 07 The Tribonacci numbers T n satisfy T 0 0, T T, and Prove that for any integer n T n T n + T n + T n 3, for n 3 ( T k T k T k Editor s Remark Most solvers first established the identity T n + T n n T k, usually by induction, and used it to complete the proof For instance, one may use induction again, and the inductive step may proceed as follows: ( n+ T k T k T k T k + T n+ T n+ ( T k + (T n+ + T n + T n T n+ ( ( T k + Tn+ + T k T n+ [( ] T k + T n+ ( n+ T k One solver submitted a bijective proof, using the polyominoes approach popularized by Arthur Benjamin and Jennifer Quinn Solution by Charles Burnette, Academia Sinica, Taipei, Taiwan Let t n denote the number of ways to tile a n rectangular board with monominoes, dominoes, and trominoes, such that the leftmost piece placed on the board is a monomino It is easy to see that t 0 0, t, t, and t n t n + t n + t n 3 for n 3; so that t n T n for all n We now count the number of ways to individually tile two n rectangular boards, one atop the other, with (horizontal monominoes, dominoes, and trominoes, such that (i the leftmost piece placed on each board is a monomino, (ii both tilings end on the right side with an identical (possibly empty sequence of dominoes, and (iii the immediate cell to the left of this sequence of dominoes on the bottom board is covered by a monomino We shall call such a tiling proper 8 VOLUME 56, NUMBER

7 ELEMENTARY PROBLEMS AND SOLUTIONS Below is an example of a proper tiling with n 0 (black dots connected by line segments mark the length of a tile First, we consider the location of the last monomino on the bottom board Since both boards are of even length and end in sequences of dominoes, the bottom board s last monomino must occupy an even-numbered cell If this monomino occupies cell k, then there are T k and T k ways to properly tile the top and bottom boards, respectively There are thus n T kt k proper tilings altogether For an alternative enumeration, note that because the properly tiled boards are of even length and begin with a monomino, the top board must have at least one other odd-length tile piece If the last odd-length tile piece on the top board is a monomino occupying cell i, then, to have a proper tiling, the last monomino on the bottom board can occupy cell j with i j n In this case, the number of proper tilings satisfying this condition is i j n T i T j On the other hand, if the last odd-length tile piece on the top board is a tromino occupying cells i through i +, then, to again have a proper tiling, the last monomino on the bottom board can occupy cell j with i + j n Hence, the number of proper tilings satisfying this condition is T i T j Combining each case yields that T k T k T i T j + ( T i T j T k i j n Also solved by Brian D Beasley, Brian Bradie, Kenny B Davenport, Steve Edwards, I V Fedak, Dmitry Fleischman, Robert Frontczak, Kantaphon Kuhapatanakul, Wei-Kai Lai and John Risher (student (jointly, Carlos Montoya (student, Ángel Plaza, Raphael Schumacher (student, Albert Stadler, and the proposer Faà di Bruno to the Rescue! B-0 Proposed by Taras Goy, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine (Vol 55, May 07 Prove that t +t + +nt nn where s t + t + + t n ( n s s! t! t! t n! F t F t F n tn ( n, Solution by Kantaphon Kuhapatanakul, Kasetsart University, Bangkok, Thailand Merca [] showed the formula for the determinant of the following Toeplitz-Hessenberg matrix as MAY 08 83

8 THE FIBONACCI QUARTERLY a a a a a a 3 a a 0 0 a n a n a n 3 a a 0 a n a n a n a a t +t + +ntnn t +t + +tns ( a 0 n s s! t! t! t n! at a t a tn n Set a 0, and a i F i for i n, and using Problem B-9 (Volume 543, we obtain the result Solution by Albert Stadler, Herrliberg, Switzerland We assume that n Let f(x + x, and g(x x k x 5 x x, x < Then by the formula of Faà di Bruno for the nth derivative of a composite function, we find n! Dn (f g t! t! t n! ((D t +t + +t n f g n ( D m g tm m! We have t +t + +nt nn m D t +t + +t n f ( t +t + +t n (t + t + + t n! x0 ( + x t +t + +t n+ ( t +t + +t n (t + t + + t n!, and Dm g m! F m Therefore, x0 n! Dn (f g ( t +t + +t n (t + t + + t n! x0 t! t! t n! t +t + +nt nn x0 n m F tm m ( On the other hand, (f g(x + x x x x x + ( + x ( x implies that n! Dn (f g(x ( ( n n! n! ( + x n+ n! ( x n+ Hence, n! Dn (f g ( n ( x0 The conclusion follows by comparing ( and ( References [] M Merca, A note on the determinant of Toeplitz-Hessenberg matrix, Special Matrices (03, 0 6 Also solved by I V Fedak, and the proposer 84 VOLUME 56, NUMBER