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 email at kwong@fredoniaedu 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 February 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-94 (Corrected Proposed by D M Bătineţu-Giurgui, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania L (L +L +m+ + L (L +L +L 3 +m+ + + L n (L +L + +L n+ +m+ L n+ 3 (3L n+ m+ for any positive integers n and m B- Proposed by Hideyuki Ohtsuka, Saitama, Japan For n, prove that F 3 n + k= Fk 3 = F 3n + 76 VOLUME 55, NUMBER 3
B- Proposed by D M Bătineţu-Giurgui, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania for any positive integer n Fn 4 + n Fn F n + + k= F 4 k +F4 k+ F k F kf k+ +F k+ > F n F n+ B-3 Proposed by Ivan V Fedak, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine For every positive integer n, prove that F F 3 F5 F 7 F4n 3 F 4n > 4 F +F 5 + +F 8n+, and F F 4 F6 F 8 F4n F 4n < 4 F 3 +F 7 + +F 8n+3 B-4 Proposed by Hideyuki Ohtsuka, Saitama, Japan Given an integer m, find a closed form for the infinite sum n= F n+m F n F n+ F n+m F n+m B-5 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 {F k,n } n N and {L k,n } n N are defined recursively by u n+ = ku n + u n for n, with respective initial conditions F k,0 = 0, F k, =, and L k=0 =, L k, = k Let c be a positive integer The sequence {a n } n N is defined by a =, a = 3, and a n+ = a n +c for n n= n= ( tan Fk,c F k,an+c tan ( Lk,c L k,an+c ( = tan k ( = tan k, if c is even;, if c is odd AUGUST 07 77
THE FIBONACCI QUARTERLY SOLUTIONS Binet! Binet! Binet! B-9 Proposed by Hideyuki Ohtsuka, Saitama, Japan (Vol 543, August 06 For nonnegative integers m and n, prove that ( n ( k k k=0 Lmk L k m = L mn L n m Solution by Jason L Smith, Richland Community College, Decatur, IL The fraction in the summand can be written as L mk L k = αmk +β mk ( α m k β m k m L k = +( m L m L m The binomial theorem can be applied thus: ( n ( k Lmk ( [ ( n α k L k = ( k m k ] β m k +( k=0 m k L m L m k=0 n n = ( αm +( βm L m L m ( β m n α m n = +( L m = βmn +α mn L n m = L mn L n m Also solved by Brian Bradie, Kenny B Davenport, Steve Edwards, I V Fedak, Dmitry Fleischman, Ángel Plaza, Hemlatha Rajpurohit, David Terr, and the proposer L m Lower Hessenberg Matrix B-9 Proposed by T Goy, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine (Vol 543, August 06 Let M n be an n n matrix given for all n by F 0 0 0 0 F F 0 0 0 F 3 F F 0 0 0 M n = F n F n F n 3 F F F n F n F n F 3 F F 78 VOLUME 55, NUMBER 3
det(m n = { 0 if n is even, if n is odd Solution by Brian Bradie, Christopher Newport University, Newport, VA Note that ( F M = (F and M =, F F so that det(m = and det(m = 0 For n 3, subtracting the second column from the first column of M n leads to det(m n = 0 0 0 0 0 0 F 0 0 0 F F F 0 0 0 F n 3 F n F n 3 F F F n F n F n F 3 F F Subtraction of the third column from the second column further reduces the determinant to 0 0 0 0 0 0 0 0 0 0 F 0 F 0 0 0 det(m n = = det(m n F n 3 F n 4 F n 3 F F F n F n 3 F n F 3 F F The recurrence relation det(m n = det(m n together with det(m = and det(m = 0 yields { 0 if n is even, det(m n = if n is odd Editor s Notes Kuhapatanskul and Plaza (independently remarked that the matrix is a lower Hessenberg matrix, whose determinant is given in [] Kenny B Davenport commented that this problem is a special case of a result in [] References [] N D Cahill, J R D Errico, D A Narayan, and J Y Narayan, Fibonacci determinants, College Math J, 33 (00, 5 [] A J Macfarlan, Use of determinants to present identities involving Fibonacci and related numbers, The Fibonacci Quarterly, 48 (00, 68 76 Also solved by Jeremiah Bartz, Hsin-Yun Ching (student, Steve Edwards, I V Fedak, Dmitry Fleischman, Kantaphon Kuhapatanakul, Mithun Kumar das, Ángel Plaza, David Stone and John Hawkins (jointly, Dan Weiner, and the proposer AUGUST 07 79
THE FIBONACCI QUARTERLY Picking the Right Numbers B-93 Proposed by José Luis Diaz-Barrero, Barcelona Tech, Barcelona, Spain (Vol 543, August 06 If F,F,,F n are the square of the first n Fibonacci numbers, then find real numbers a,a,,a n satisfying a k > Fk, k n, and a k < α F n F n+ α Solution by Steve Edwards, Kennesaw State University, Marietta, GA k= Since α α = α+ α >, we can let a k be any real number satisfying Fk < a k < α Then, since n k= F k = F nf n+, we find F n F n+ a k < F n F n+ k= k= α α F k = α F n F n+ α F nf n+ = α α α F k Also solved by Brian D Beasley, Brian Bradie, Kenny B Davenport, I V Fedak, Dmitry G Fleishcman, Ángel Plaza, David Stone and John Hawkins (jointly, and the proposer Errors! B-94 Proposed by D M Bătineţu-Giurgui, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania (Vol 543, August 06 L (L +L +m+ + L (L +L +L 3 +m+ + + L n (L +L + +L n+ +m+ (L n+ m+ L m+ n+ (L n+ 3 m for any positive integers n and m Editor s Remark The right-hand side of the inequality was incorrectly stated in the original problem The correct version can be found at the beginning of the section in this issue Polygon with Generalized Fibonacci Numbers as Its Vertices B-95 Proposed by Jeremiah Bartz, Francis Marion University, Florence, SC (Vol 543, August 06 Let G i denote the generalized Fibonacci sequence given by G 0 = a, G = b, and G i = G i +G i for i 3 Let m 0 and k 0 the area A of the polygon with n 3 vertices (G m,g m+k,(g m+k,g m+3k,,(g m+(n k,g m+(n k 80 VOLUME 55, NUMBER 3
is where µ = a +ab b µ F k ( Fk(n (n F k Solution by Ivan V Fedak, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine It is known [, Theorem 333] that the area with vertices (G n,g n+r, (G n+p,g n+p+r, and (G n+q,g n+q+r is independent of n, and equals ( µfr ( p F q p +F p F q For n = m, r = k, p = ks, where s n, and q = p+k, we obtain the area ( µfk Fk +F ks F k(s+ = µ F ( k Fk(s+ F ks F k Therefore, A = n µ F ( k Fk(s+ F ks F k s= = µ F ( k Fk(n F k (n F k = µ F k( Fk(n (n F k References [] T Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley, New York, 00 Also solved by the proposer AUGUST 07 8