ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS EDITED BY HARRIS KWONG Please submit solutios ad problem proposals to Dr Harris Kwog, Departmet of Mathematical Scieces, SUNY Fredoia, Fredoia, NY, 4063, or by at If you wish to have receipt of your submissio acowledged by mail, please iclude a selfaddressed, stamped evelope Each problem or solutio should be typed o separate sheets Solutios to problems i this issue must be received by August, 208 If a problem is ot origial, the proposer should iform the Problem Editor of the history of the problem A problem should ot be submitted elsewhere while it is uder cosideratio for publicatio i this Joural Solvers are ased to iclude refereces rather tha quotig well-ow results The cotet of the problem sectios of The Fiboacci Quarterly are all available o the web free of charge at wwwfqmathca/ BASIC FORMULAS The Fiboacci umbers F ad the Lucas umbers L satisfy F 2 F F, F 0 0, F ; L 2 L L, L 0 2, L Also, α ( )/2, β ( )/2, F (α β )/, ad L α β PROBLEMS PROPOSED IN THIS ISSUE B-22 Proposed by José Luis Díaz-Barrero, Techical Uiversity of Cataloia (Barceloa Tech), Barceloa, Spai For ay positive iteger, show that 4 F L 4F 2 F (F L ) 2 F 2 L F 2 F2 2 is a perfect square, ad fid its value B-222 Proposed by Key B Daveport, Dallas, PA Let H deote the th harmoic umber Prove that H F l 6 l α H 2 L, ad 2 (l 2) 2 4(l α) 2 2 FEBRUARY

2 THE FIBONACCI QUARTERLY B-223 Proposed by Iva V Feda, Vasyl Stefay Precarpathia Natioal Uiversity, Ivao-Fraivs, Uraie For all positive itegers ad a, prove that F (F a F 2 a F 2 a ) 0 B-224 Proposed by Hideyui Ohtsua, Saitama, Japa For ay positive iteger, prove that F F 2, ad L L 2 2 B-22 Proposed by Jatha Austi, Salisbury Uiversity, Salisbury, MD Costruct a sequece {M } of 3 3 matrices with positive etries that satisfy the followig coditios: (A) M is the product of ozero Fiboacci umbers (B) The determiat of ay 2 2 submatrix of M is a Fiboacci umber or the product of ozero Fiboacci umbers (C) lim M / M 2α SOLUTIONS Cauchy-Schwarz or Bergström Agai! B-20 Proposed by Iva V Feda, Vasyl Stefayc Precarpathia Natioal Uiversity, Ivao-Fraivs, Uraie (Vol, February 207) If a, b, c > 0, the prove that, for ay positive iteger, af bf cf 2 al bl cl 2 a2 b 2, af bf bf af F 2 a2 b 2, al bl bl al L 2 bf cf af 2 bl cl al 2 c 3 a2 b 2 c 2, cf af bf 2 2F 2 c 3 a2 b 2 c 2 cl al bl 2 2L 2 Solutio by Bria Bradie, Christopher Newport Uiversity, Newport News, VA Let a, b, c, x, y, z be positive real umbers By the Cauchy-Schwarz iequality, ax by b3 bx cz a 4 a 2 x aby b 4 b 2 x aby (a 2 b 2 ) 2 (a 2 b 2 )x 2aby 82 VOLUME 6, NUMBER

3 ELEMENTARY PROBLEMS AND SOLUTIONS Now, by the arithmetic mea - geometric mea iequality, 2ab a 2 b 2, so ax by b3 bx cz (a 2 b 2 ) 2 (a 2 b 2 )x (a 2 b 2 )y a2 b 2 x y () With x F ad y F, () becomes a2 b 2 a2 b 2 af bf bf af F F F 2 With x L ad y L, () becomes a2 b 2 a2 b 2 al bl bl al L L L 2 Next, by the Cauchy-Schwarz iequality, ax by cz bx cy az c 3 cx ay bz a 4 a 2 x aby caz b 4 b 2 x bcy abz c 4 c 2 x cay bcz (a 2 b 2 c 2 ) 2 (a 2 b 2 c 2 )x (ab bc ca)(y z) Now, the iequality (a b) 2 (b c) 2 (c a) 2 0 is equivalet to so ab bc ca a 2 b 2 c 2, ax by cz bx cy az c 3 cx ay bz (a 2 b 2 c 2 ) 2 (a 2 b 2 c 2 )x (a 2 b 2 c 2 )(y z) a2 b 2 c 2 x y z (2) With x F, y F, ad z F 2, (2) becomes c 3 af bf cf 2 bf cf af 2 cf af bf 2 a 2 b 2 c 2 a2 b 2 c 2 F F F 2 2F 2 With x L, y L, ad z L 2, (2) becomes c 3 al bl cl 2 bl cl al 2 cl al bl 2 a 2 b 2 c 2 a2 b 2 c 2 L L L 2 2L 2 Editor s Note: Ricardo used Bergström iequality to derive () ad (2) Also solved by Dmitry Fleischma, Hideyui Ohtsua, Ágel Plaza, Hery Ricardo, Nicuşor Zlota, ad the proposer FEBRUARY

4 THE FIBONACCI QUARTERLY Root ad Ratio Tests B-202 Proposed by D M Bătieţu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia; Neculai Staciu, George Emil Palade School, Buzău, Romaia; ad Gabriel Tica, Mihai Viteazul Natioal College, Băileşti, Dolj, Romaia (Vol, February 207) a Let (a ) be a positive real sequece such that lim 2 a Evaluate a ( ) ( lim a F ( )! a F! ad lim a L ( )! a L! ) Solutio by the proposers We claim that both limits equal to aα/e Give a ifiite sequece (b ), it is ow that if lim b /b L, the lim b L Apply this to c a F /(! ) We fid lim c c lim Thus, lim c aα/e as well Defie u a F! a lim ( )! ( ) a F 2 F 2 aα a F e a F ( )!! c a F c, such that a F ( )! a F! a F! ( u ) c (u ) It suffices to show that lim (u ) Note that lim u, ad c lim u lim c e c Therefore, lim (u u ) lim l u l e l u The proof of the other limit is similar, ad is omitted here Editor s Note: Plaza oted that the iequalities follow from a result obtaied by the first two proposers i [], ad Ohtsua used a result from [2] to derive the iequalities directly 84 VOLUME 6, NUMBER

5 ELEMENTARY PROBLEMS AND SOLUTIONS Refereces [] D M Bătieţu-Giurgiu ad N Staciu, New methods for calculatios of some limits, The Teachig of Mathematics, 6(2) (203), [2] Gh Toader, Lalescu sequeces, Publiacije Eletrotechičog faulteta Uiverziteta u Beogradu, Serija Matematia i fizia, 9 (998), 9 28 Also solved by I V Feda, Dmitry Fleishcma, Hamza Mahmood (studet), Soumitra Madal, Hideyui Ohtsua, Ágel Plaza, ad Raphael Schumacher (studet) Fiboacci Numbers with Fiboacci Numbers as Subscripts B-203 Proposed by Hideyui Ohtsua, Saitama, Japa (Vol, February 207) Prove that, for ay positive iteger, (i) F F3 F F3 F F3 2 3 F 2F ; (ii) L L3 L L3 L L ( ) L L 2L Solutio by Jaroslav Seibert, Uiversity of Pardubice, Czech Republic Usig the Biet s formula for the Fiboacci umbers, we fid ( α F 3 β F ) ( 3 α F 3 β F ) ( 3 α F 3 2 β F ) 3 2 F F3 F F3 F F3 2 [ α 2F 3 (αβ) F 3 2 α 2F 3 (αβ) F 3 α 2F 3 2 (αβ) F 3 (αβ) F 3 (αβ) F 3 β 2F 3 2 (αβ) F 3 2 β 2F 3 β 2F ] 3 Sice αβ, ad F 3 ad F 3 2 are both odd for ay iteger, we have (αβ) F 3 (αβ) F 3 2 Thus, F F3 F F3 F F3 2 ( α 2F 3 β 2F 3 α2f 3 β 2F 3 α2f 3 2 β 2F ) 3 2 F2F3 F 2F3 F 2F3 2 Fially, F F3 F F3 F F3 2 which proves (i) ( F2F3 F 2F3 F 2F3 2 ) 3 F 2F, FEBRUARY 208 8

6 THE FIBONACCI QUARTERLY The proof of (ii) proceeds i a similar maer Usig the Biet s formula for the Lucas umbers, we fid L L3 L L3 L L3 2 ( α L 3 β L 3 ) ( α L 3 β L 3 ) ( α L 3 2 β L 3 2 ) α 2L 3 (αβ) L 3 2 α 2L 3 (αβ) L 3 α 2L 3 2 (αβ) L 3 (αβ) L 3 (αβ) L 3 β 2L 3 2 (αβ) L 3 2 β 2L 3 β 2L 3 It is ow that L 3 ad L 3 2 are both odd, ad L 3 is eve for ay iteger Hece, (αβ) L 3 (αβ) L 3 2, ad (αβ) L 3 Thus, L L3 L L3 L L3 2 ( α 2L 3 β 2L 3 ) ( α 2L 3 β 2L 3 ) ( α 2L 3 2 β 2L 3 2 ) 2 L 2L3 L 2L3 L 2L3 2 2 ( ) L 3 L 2L3 ( ) L 3 L 2L3 ( ) L 3 2 L 2L3 2 2, which proves that L L3 L L3 L L ( ) L L 2L Editor s Note: Plaza quoted the geeral formulas for the products F x F x2 F x3 ad L x L x2 L x3 i [2], ad Daveport applied the followig symmetric idetities from []: F x F y F z F xyz ( ) x F xyz ( ) y F x yz ( ) z F xy z, L x L y L z L xyz ( ) x L xyz ( ) y L x yz ( ) z L xy z Refereces [] P S Brucma, Solutio to Problem B-890, The Fiboacci Quarterly, 38 (2000), [2] H H Fers, Products of Fiboacci ad Lucas umbers, The Fiboacci Quarterly, 7 (969), 3 Also solved by Bria Bradie, Key B Daveport, I V Feda, Dmitry Fleischma, Ágel Plaza, Raphael Schumacher (studet), ad the proposer A Double Biomial Sum B-204 Proposed by Steve Edwards, Keesaw State Uiversity, Marietta, GA (Vol, February 207) For o-egative itegers, express j j i j A 2 2j j i j0 i0 i terms of Fiboacci umbers ad B j0 2 2j j j i j j i i0 Solutio by Hideyui Ohtsua, Saitama, Japa We use the well-ow idetity /2 0 F 86 VOLUME 6, NUMBER

7 ELEMENTARY PROBLEMS AND SOLUTIONS We have ad A B j0 j0 j0 j0 j0 j0 j0 j0 j0 j 2 2j i0 j 2 2j i0 2 2j 2 2j ( j j ( j j j 2 2j i0 j 2 2j i0 ( j i)! ( j)! (2j i)! ( j)! ( j)! (2j)! ) j i0 ( j 2 2j j ( j 2 2j j 2j i ) 2 2j ( j)! i! ( j i)! (2j)! i! (2j i)! j 2j 2 2j j j0 i0 2 F 2, 0 ( j i)! ( j )! (2j i )! ( j)! ( j )! (2j )! ) j 2j i i0 ) 2j i0 2j j 2 2j 2 2j j i 0 ( j)! i! ( j i)! (2j )! i! (2j i )! 2j i 2 F 2 Also solved by Bria Bradie, I V Fade, Dmitry Fleischma, Jaroslav Seibert, ad the proposer Power-Mea ad Jese s Iequalities B-20 Proposed by D M Bătieţu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia; ad Neculai Staciu, George Emil Palade School, Buzău, Romaia (Vol, February 207) Prove that for ay positive itegers ad m m F 2m F m F m FEBRUARY

8 THE FIBONACCI QUARTERLY Solutio by Ágel Plaza, Uiversidad de Las Palmas de Gra Caaria, Spai It is eough to apply the power-mea arithmetic mea iequality to the sequece ( F 2 ) as follows: m F 2m F 2 F F It follows that m F 2m F m F m Solutio 2 by Hery Ricardo, New Yor Math Circle, Purchase, NY, Notig that, for ay positive iteger m, the fuctio f(x) x m is covex o the iterval (0, ), ad that F 2 F F, we use Jese s iequality to coclude that f ( F 2 ) f F 2, or m F 2m (F F ) m F m F m Also solved by Maria Aristizabal (studet), Bria Bradie, Key B Daveport, I V Feda, Dmitry Fleischma, Wei-Kai Lai, Soumitra Madal, Hideyui Ohtsua, Nicuşor Zlota, ad the proposer 88 VOLUME 6, NUMBER