Edited by Russ Euler and Jawad Sadek

Transcription

1 Edited by Russ Euler and Jawad Sade Please submit all new problem proposals and corresponding solutions to the Problems Editor, DR RUSS EULER, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 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 selfaddressed, stamped envelope Each problem and solution should be typed on separate sheets Solutions to problems in this issue must be received by February 5, 2006 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 BASIC FORMULAS The Fibonacci numbers F n and the Lucas numbers L n satisfy F n+2 = F n+ + F n, F 0 = 0, F = ; L n+2 = L n+ + L n, L 0 = 2, L = Also, α = ( + 5)/2, β = ( 5)/2, F n = (α n β n )/ 5, and L n = α n + β n PROBLEMS PROPOSED IN THIS ISSUE B-00 Proposed by Paul S Brucman, Canada (a) Let θ(a, b) = sin {(a 2 b 2 )/(a 2 + b 2 )} denote one of the acute angles of a Pythagorean triangle, where a and b are integers with a > b > 0 Given θ(a, b) and θ(c, d), show that there exists θ(e, f), where e and f are functions of a, b, c and d, such that θ(a, b) + θ(c, d) + θ(e, f) = π/2 (b) As a special case, prove the following identity: sin {F n F n+2 /F 2n+ } + sin {L n L n+2 /5F 2n+ } + sin {3/5} = π/2 277

2 B-002 Proposed by H-J Seiffert, Berlin, Germany Prove the identity ( ( ) F+ = F ) ( α = β + 2 = F F + ) B-003 Proposed by Paul S Brucman, Canada Prove the identity F n+2 L n+ L n F n + L n+2 F n+ F n L n = 2(F 2n+ ) 2 B-004 Proposed by José Luis Díaz-Barrero, Barcelona, Spain Let n be a positive integer Calculate { lim n F n ( n n i= = F i+ n = F i n i= )} B-005 Proposed by H-J Seiffert, Berlin, Germany Prove that, for all integers and n with 0 n, 2n 2 j=0 ( ) j ( 2n + j )( 2n j ) F j = 0 278

3 SOLUTIONS A Closed Form for a Fibonacci Sum B-986 Proposed by Br J Mahon, Australia Prove i=2 F 2i 2 F 2i 3 (F 2 2i ) ( F 2 2i+2 ) = 8 + F 2nF 2n+2 3 ( F 2 2n+2 ) Solution by Steve Edwards, Southern Polytechnic State University, Marietta, GA The formula is easily verified for small n, so we assume it to be true for values through n and proceed by induction: n+ i=2 F 2i 2 F 2i 3(F 2 2i )(F 2 2i+2 ) = 8 + F 2nF 2n+2 3(F 2 2n+2 ) + F 2n F 2n+2 3(F 2 2n+2 )(F 2 2n+4 ) = 8 + F 2nF 2n+2 (F 2 2n+4 ) + F 2n F 2n+2 3(F 2 2n+2 )(F 2 2n+4 ) = 8 + F 2n F 2n+2 F 2 2n+4 3(F 2 2n+2 )(F 2 2n+4 ) But since F 2n F 2n+4 = (F 2 2n+2 ) (see eg []), this equals 8 + F 2n+2F 2n+4 3(F 2 2n+4 ) Reference Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, 200 All solutions were more or less similar to the featured one Also solved by Brian D Beasley, Paul S Brucman, Kenneth B Davenport, Ovidiu Furdui, George C Greubel, Russell J Hendel, Harris Kwong, Hradec Králové, H-J Seiffert, and the proposer Another Toss B-987 Proposed by MN Deshpande and JP Shiwalar, Nagpur, India An unbiased coin is tossed n times Let A be the event that no two successive heads occur Show P r(a) = (F n+2 )/2 n Solution by Maitland Rose and Michael Becer, University of South Carolina, Sumter P r(a) =(Number of sequences of coins for which no two heads are together)/2 n Fibonacci s tree diagram for his rabbit model applies directly to our present case If A (adults) is replaced by T (tails) and B (babies) is replaced by H (heads) we may interpret no two successive heads as a baby does not produce a baby 279

4 If we start with a T, then for n tosses the number of outcomes with no two successive heads is F n+ If we start with a H, then for n tosses the number of outcomes with no two successive heads is F n Therefore P r(a) = F n++f n 2 = F n+2 n 2 n Comment As most of the solvers pointed out, this problem is very well nown H-J Seiffert mentions that a proof was given recently in The Mathematical Gazette, 88, November 2004; by M Griffiths He also mentions that the problem is very similar to problem B-688 of The Fibonacci Quarterly Pentti Hauanen mentions that this problem occurs as an example or as an exercise, more or less in the same form, in most textboos on combinatorics and discrete mathematics (see R Grimaldi, example 02 or T Koshy, Chapter 4 to name a few) Also solved by Charles Ashbacher, Brian Beasley, Paul S Brucman, Steve Edwards, George C Greubel, Pentti Hauanen, Russell Hendel, Hradec Králové, Harris Kwong, Kathleen E Lewis, Graham Lord, William Moser, H-J Seiffert, James Sellers, and the proposer It s All in the Parity B-988 Proposed by Steve Edwards, Southern Polytechnic State University, Marietta, GA Show that for an odd integer n and any integer, 5(F 2 F n 2 ) + 4( ) is the product of two Lucas numbers, while if n is an even integer, then F 2 F n 2 is the product of two Fibonacci numbers Solution by Brian D Beasley, Presbyterian College, Clinton, SC Using αβ =, we note the following: α n β 2 n = (αβ) n β 2 2n = ( ) n β 2 2n ; β n α 2 n = (αβ) n α 2 2n = ( ) n α 2 2n Then for even n, using the above results along with the observation that ( ) = ( ) n, we obtain F 2 F 2 n = 5 [ α 2 2( ) + β 2 α 2 2n + 2( ) n β 2 2n] = 5 (αn β n )(α 2 n β 2 n ) = F n F 2 n Similarly, for odd n, we obtain F 2 F 2 n = 5 [ α 2 + β 2 α 2 2n β 2 2n 4( ) ] 280

5 and thus 5(F 2 F 2 n) + 4( ) = α 2 + β 2 α 2 2n β 2 2n = (α n + β n )(α 2 n + β 2 n ) = L n L 2 n Also solved by Paul S Brucman, Charles K Coo, Ovidiu Furdui, George C Greubel, Russell J Hendel, Hradec Králové, Harris Kwong, William Moser, Maitland Rose, H-J Seiffert, James Sellers, and the proposer A Lower Bound for a Fibonacci Sum B-989 Proposed by José Luis Díaz-Barrero, Universidad Politécnica de Catalunya, Barcelona, Spain Let n be a positive integer ( Prove that ) n e n F 2n n F C(n, ) = Solution by H-J Seiffert, Berlin, Germany The Cauchy-Schwarz Inequality gives ( n ) ( n ) F C(n, ) n 2 F C(n, ) = = Since (see, for example, P Hauanen On a Binomial Sum for the Fibonacci and Related Numbers The Fibonacci Quarterly 344 (996): 326-3, () and (2)) we obtain F C(n, ) = F 2n, = = F C(n, ) n2 F 2n The nown inequality e x + x, x R, with x = F 2n /n implies n F 2n e F 2n/n The stated inequality follows Also solved by Paul S Brucman, Steve Edwards, Ovidiu Furdui, George C Greubel, Russell J Hendel, Hradéc Králové, Harris Kwong, and the proposer 28

6 Two Binomial-Type Identities B-990 Proposed by Mario Catalani, University of Torino, Torino, Italy Let F n (x, y) and L n (x, y) be the bivariate Fibonacci and Lucas polynomials, defined, respectively, by F n (x, y) = xf n (x, y)+yf n 2 (x, y), F 0 (x, y) = 0, F (x, y) = and L n (x, y) = xl n (x, y)+yl n 2 (x, y), L 0 (x, y) = 2, L (x, y) = x Assume x 2 +4y 0 Prove the following identities A) B) =0 F (x, y)f n (x, y) = =0 x 2 + 4y (2n L n (x, y) 2x n ), L (x, y)l n (x, y) = 2 n L n (x, y) + 2x n Solution by Paul S Brucman, Sointula, BC, and Harris Kwong, SUNY Fredonia, Fredonia, NY (independently) Let α = α(x, y) = (x + x 2 + 4y)/2 and β = β(x, y) = (x x 2 + 4y)/2 We can derive the Binet s formulas for F n (x, y) and L n (x, y) in the same manner they were derived for the ordinary Fibonacci and Lucas numbers We find F n (x, y) = αn β n α β = αn β n x2 + 4y and L n (x, y) = α n + β n A) (x 2 + 4y) =0 F (x, y)f n (x, y) = B) In a similar manner, we also find =0 = =0 =0 (α β )(α n β n ) [(α n + β n ) (α β n + α n β )] = (α n + β n ) =0 = 2 n L n (x, y) 2x n 2(α + β) n L (x, y)l n (x, y) = 2 n L n (x, y) + 2x n Also solved by Kenneth B Davenport, Ovidiu Furdui, George C Greubel, Hradec Králové, H-J Seiffert, and the proposer We wish to belatedly acnowledge the solution to problem B-976 by Scott Brown 282