ELEMENTARY PROBLEMS AND SOLUTIONS

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1 EDITED BY RUSS EULER AND JAWAD SADEK Please submit all new problem proposals and their solutions to the Problems Editor, DR RUSS EULER, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468, or by at All solutions to others proposals must be submitted to the Solutions Editor, DR JAWAD SADEK, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO If you wish to have receipt of your submission acnowledged, please include a self-addressed, stamped envelope Each problem and solution should be typed on separate sheets Solutions to problems in this issue must be received by November 15, 015 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 ased to include references rather than quoting well-nown 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+1 +F n, F 0 = 0, F 1 = 1; L n+ = L n+1 +L n, L 0 =, L 1 = 1 Also, α = (1+ 5)/, β = (1 5)/, F n = (α n β n )/ 5, and L n = α n +β n PROBLEMS PROPOSED IN THIS ISSUE B-1166 Proposed by Hideyui Ohtsua, Saitama, Japan Prove that F 1 L =1 = B-1167 Proposed by Atara Shrii and Opher Liba, Oranim College of Education, Israel Find the area of the polygon with vertices (F 1,F ),(F 3,F 4 ),(F 5,F 6 ),,(F n 1,F n ) for n VOLUME 53, NUMBER

2 B-1168 Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Let n be a nonnegative integer Find the minimum value of F n F n +3L n (1+F n ) + F n L n F n L n +3(F n +L n ) + L n L n +3F n (1+L n ) B-1169 Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain Let p be a positive integer Prove that for any integer n > 1 F n+ F p+1 p n+1 Fp+1 n F p n p +F p n+1 (p+1)f n 1 B-1170 Proposed by D M Bătineţu Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania Let a > 0, b > 0, and c 0 Find (i) lim (n+1) n+1 (n+1)!afn+1 c (ii) lim (n+1) n+1 (n+1)!al c n+1 n n n!bf c n, n n n!bl c n SOLUTIONS It Can Be Sharpened B-1146 Proposed by Titu Zvonaru, Comăneşti, Romania (Vol 5, May 014) Prove that + 5F n 1 for all integers n 1 Solution by Brian D Beasley, Department of Mathematics, Presbyterian College, Clinton, SC We claim that for any positive real number c with c < α 3, Fn+ c Fn 1 for sufficiently large n In particular, Fn+ 9F n 1 for n 1 Fix a real number c with 0 < c < α 3 = + 5 Since F n+ cf n 1 trivially for n = 1, we move on to consider n > 1 Then F n+ cf n 1 if and only if α n 1 (α 3 c) β n 1 (β 3 c) MAY

3 THE FIBONACCI QUARTERLY If n is odd, then this inequality holds trivially, as its left side is positive while its right side is negative If n is even, then this inequality holds if and only if d n 1 e, where d = α/( β) = α+1 > 1 and e = (c β 3 )/(α 3 c) > 0 Thus the inequality holds if and only if n log(e) log(d) +1, so we conclude that F n+ cf n 1 for sufficiently large n In particular, for c = 3, we have F n+ cf n 1 for n Addendum We note that we may let c be arbitrarily close to α 3 For example, if c = 4, then F n+ cf n 1 for n 6; if c = 43, then F n+ cf n 1 for n 8; etc Also solved by Carlos Alirio Rico Acevedo, Gurdial Arora, Charles K Coo, Kenneth B Davenport, Pedro Fernando Fernández Espinoza, Dmitry Fleischman, Ralph P Grimaldi, Russell Jay Hendel, (Tia Herring, Gary Knight, Shane Latchman, Ludwig Murillo, Zibusiso Ndimande, and Eyob Tareegn; students) (jointly), Wei-Kai Lai, Kathleen E Lewis, Sydney Mars and Leah Seader (jointly), Reiner Martin, Grey Martinsen and Pantelimon Stănică (jointly), Ángel Plaza, Jason L Smith, and the proposer Close This Sum B-1147 Proposed by Hideyui Ohtsua, Saitama, Japan (Vol 5, May 014) Find a closed form expression for 0 a,b,c n a+b+c=n F a F b F c a!b!c! Solution by Harris Kwong, SUNY Fredonia, Fredonia, NY The sum is in the form of a convolution, its generating function is Hence, G(x) = n 0 0 a,b,c, n a+b+c=n F a F b F c a!b!c! xn = F a a 0 G(x) = α β α β x! 0 3 a! xa b 0 F b b! xb ( e αx e βx ) 3 = α β Since α+β = α+1 = α, and α+β = β +1 = β, we find (α β) 3 G(x) = e 3αx e (α+β)x +3e (α+β)x e 3βx = e 3αx e 3βx 3(e αx e βx ) c 0 F c c! xc 18 VOLUME 53, NUMBER

4 After comparing the coefficients of x n from both sides of this equation, we determine that 0 a,b,c, n a+b+c=n F a F b F c a!b!c! = (3α)n (3β) n 3(α n β n ) (α β) 3 n! = 3n F n 3F n 5n! Also solved by Dmitry Fleischman, (Sydney Mars, Leah Seader, and Michelle Manin; students) (jointly), Reiner Martin, Ángel Plaza and Sergio Falcón (jointly), and the proposer The Exact Value of an Infinite Series B-1148 Proposed by Hideyui Ohtsua, Saitama, Japan (Vol 5, May 014) Prove that =1 F 1 L +1 = 1 5 Solution by students Tia Herring, Gary Knight, Shane Latchman, Ludwig Murillo, Zibusiso Ndimande and Eyob Tareegn, jointly, Benedict College, Columbia, SC We will first prove by induction on n that n F 1 L +1 = L n +1 If n = 1, then the equality becomes L n +1 Then n+1 =1 =1 F n 1 L +1 = F 1 L +1 = F L +1, which is true Assume that n F 1 =1 L +1 = =1 F 1 L +1 + L n 1 +1 = L n +1 + = L n 1 L n 1 L n 1 +1 L n +1 + L n+1 +1 = (L n 1) (L n) 1 + L n+1 +1 L n = L n+1 +( 1) n 1 + L n 1 +1 = L n L n 1 +1 = +1 L n 1 +1 which proves that the inequality holds for n + 1 In the previous string of equalities we used the following identities (L n ) = L n + ( 1) n and L n F n = F n, [1, pp 97, 80] Now, MAY

5 THE FIBONACCI QUARTERLY F n+1 since lim = α, we have that lim F n α+ 1 α = 5 and therefore, L n F n = lim F n+1 +F n 1 F n ( ) F n+1 = lim + 1 F F n n = F n 1 =1 F 1 L +1 = lim n =1 F 1 L +1 = lim L n +1 = lim 1 L n + 1 = 1 5 References [1] T Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, Inc, 001 Also solved by Russell Jay Hendel, Reiner Martin, Ángel Plaza, and the proposer Inequalities with -Fibonacci and -Lucas Sequences B-1149 Proposed by Ángel Plaza and Sergio Falcón, Universidad de Las Palmas de Gran Canaria, Spain (Vol 5, May 014) For any positive integer number, the -Fibonacci and -Lucas sequences, {F,n } n N and {L,n } n N, both are defined recurrently by u n+1 = u n +u n 1 for n 1, with respective initial conditions F,0 = 0; F,1 = 1 and L,0 = ; L,1 = Prove that n 1 (F,i F,i F,i+1 +F,i+1 ) +(F,n F,n F,1 +F,1 ) F,nF,n+1, (1) n 1 (L,i L,i L,i+1 +L,i+1 ) +(L,n L,n L,1 +L,1 ) L,nL,n+1, () for any positive integer n Solution by the proposers Both inequalities are consequence that for any positive real numbers x, y, x+y and hence, x+y x xy + +y Then the left-hand side of (1), LHS is LHS F,1 +F, + F, +F,3 xy + x +y + + F,n +F,1, = n F,i = F,nF,n+1, 184 VOLUME 53, NUMBER

6 where the last identity may be proved easily by induction For n = 1 trivially, 1 = 1 = 1 Assume that the identity holds for n 1 Then n n 1 F,i = F,i +F,n = F,n 1F,n +F,n = F,n 1F,n +F,n = F,n(F,n 1 +F,n ) = F,nF,n+1 Inequality () is proved in the same way by now using Also solved by Dmitry Fleischman n L,i = L,nL,n+1 Maximum of a Lucas Fibonacci Ratio B-1150 Proposed by Hideyui Ohtsua, Saitama, Japan (Vol 5, May 014) Let n be a positive integer For any positive integers r 1,r,,r n, find the maximum value of L r1 +r + +r n F r1 F r F rn as a function of n Solution by Reiner Martin, 6581 Bod Soden-Nevenbdin, Germany We will show that the maximum value is L n, which is attained by r 1 = r = = r n = Note that if some r i is equal to 1, then we can replace it by to get a larger value for the fraction, as F 1 = F and as the Lucas numbers are increasing for positive indices Thus, we can assume that r i for all i = 1,,,n We proceed by showing that any such fraction does not decrease when we replace any r i by As the expression in question is symmetric in r i,r,,r n it is enough to show that L r1 +r + +r n L +r + +r n F r1 F r F rn F F r F rn for all r i,r,,r n Writing r = r + +r n, this is equivalent to L r1 +rf L +r F r1 But equation (18) in S Vajda, Fibonacci and Lucas Numbers and the Golden Section, Halsted Press (1989) implies that L +r F r1 L r1 +rf = L r F r1 L r1 +r F 0 = L r F r1 0 Also solved by Dmitry Fleischman, Russell Jay Hendel, and the proposer MAY

ELEMENTARY PROBLEMS AND SOLUTIONS

ELEMENTARY PROBLEMS AND SOLUTIONS EDITED BY RUSS EULER AND JAWAD SADEK Please submit all new problem proposals their solutions to the Problems Editor, DR. RUSS EULER, Department of Mathematics Statistics, Northwest Missouri State University,