ELEMENTARY PROBLEMS AND SOLUTIONS

Transcription

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 acnowledged 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 August 5, 09. 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 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-756 Proposed by Stanley Rabinowitz, Chelmsford, MA. (Vol. 3., February 994) Find a formula expressing the Pell number P n in terms of Fibonacci and/or Lucas numbers. Editor s Note: This is an old problem from 994 that was proposed by the former Problem Section Editor. At that time, no relatively simple and elegant solutions were received, so the editor left the problem open. At its 5th anniversary, we have revived the problem, and invite the readers to solve it. B-4 Proposed by Ivan V. Feda, Vasyl Stefany Precarpathian National University, Ivano-Franivs, Uraine. For all positive integers n, prove that F n+ + F n+ F n + >. L n+ L n+ L n+ + F n+ FEBRUARY 09 8

2 THE FIBONACCI QUARTERLY B-4 Proposed by Hideyui Ohtsua, Saitama, Japan. Let r, r,..., r n be positive even integers. Prove that n F ɛ r + +ɛ nr n 0, and L ɛ r + +ɛ nr n L r. ɛ,,ɛ n {,} ɛ,,ɛ n {,} B-43 Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain. For any positive integer, the -Fibonacci numbers are defined recursively by F,0 0, F,, and F,n F, + F,n for n. Prove that n ( F,m ) n+ n+ n + 4 (. + 4 ) m0 B-44 Proposed by Robert Frontcza, Landesban Baden-Württemberg, Stuttgart, Germany. Let n be an integer. Prove the following identities for the Fibonacci and Lucas numbers: n a) (F F j ) nf n F n+ (F n+ ). b) c) j+ n j+ n j+ (L L j ) n(l n L n+ ) (L n+ 3). (F L j F j L ) { F n F n+ (L n L n+ ) (F n+ L n ), F n F n+ (L n L n+ ) Fn+ L n, B-45 Proposed by Kenny B. Davenport, Dallas, PA. Show that, for any positive integer n, n L 3 5 [(n + 3)L 3 L 3n ] + 49 (n + )L 3 L 3 4 n. Determinant of a Symmetric Matrix if n is even, if n is odd. B- Proposed by José Luis Díaz-Barrero, Technical University of Catalonia (Barcelona Tech), Barcelona Spain. For any positive integer n, show that 4 F n L n 54F n F n (F n+ + L n ) F n L n F n Fn+ is a perfect square, and find its value. 8 VOLUME 57, NUMBER

3 ELEMENTARY PROBLEMS AND SOLUTIONS Composite solution by I. V. Feda, Vasyl Stefany Precarpathian National University. Ivano-Franivs, Uraine, and the editor. and Let x F and y F n+, so that Then, F n y x, L n y + x, F n+ y x, F n+ + L n y + x, 4 F n L n F n (F n+ + L n ) F n L n F n Fn+ F n F n L n (y x)(y + x) y x. 4 y x y + x y x (y + x) y x y + x y x (y x) 4(y + x) (y x) (y x ) (y + x) (y + x) (y x) (y x), which can be simplified to [((y x) (y + x) ] (y + x) + [(y + x) (y x) ] (y x) (y x ) (7y 0xy + x )(y + x) + (7y + 0xy + x )(y x) (y x ) (7y + x )(4y + x ) 80x y (y x ) 54y (y x ) 54F n+f n. Thus, the value of the given expression is Fn+ for any positive inteber n. Also solved by Brian D. Beasley, Dmitry Fleischman, G. C. Gruebel, Stacy M. Hartz (student), Wei-Kai Lai, Ehren Metcalfe, Kambiz Moghaddamfar (student), Raphael Schumacher (student), Jaroslav Seibert, Jason L. Smith, Albert Stadler, Nicuşor Zlota, and the proposer. The Generating Function for Harmonic Numbers B- Proposed by Kenny B. Davenport, Dallas, PA. Let H n denote the nth harmonic number. Prove that H F n ln 6 ln α H n n L n, and 5 n n (ln ) + 4(ln α). n Solution by Amanda M. Andrews and Samantha L. Zimmerman (students), California University of Pennsylvania, California, PA (jointly). We will deduce the results for the generalized Fibonacci sequence {G n } n N defined by G a, G b, and G n G + G n, for n 3. The generating function for H n is nown to be H n x n ( ) x ln, x (, ). x n FEBRUARY n

4 THE FIBONACCI QUARTERLY Now, after integrating this power series over [0, x] for x (, ), we obtain H x n n ln ( x), x (, ). We also use [, p. ] n G n cαn dβ n 5, n Z, where c a + (a b)β, and d a + (a b)α. Since α, β (, ), we find H G n n n c H ( α ) n d ( H 5 β n 5 n n n n c ) ). 5 ln ( α It is easy to verify that α, and β α β n H G n n n 5 d 5 ln ( β ) n α. Therefore, [ ( ) ( )] α c ln 5 α d ln [ c (ln + ln α) d (ln ln α) ] 5 ( ) (c d) [(ln ) + 4(ln α) ] + 4(c + d) ln ln α. For Fibonacci numbers, we have a b ; hence, c d, and H F n n n ln 6 ln α (8 ln ln α). 5 5 n For Lucas numbers, we have a and b 3; hence, c d 5, which leads to c + d 0, and c d 5. Thus, H L n n n (ln ) + 4(ln α). n References [] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, New Yor, 00. Also solved by Khristo N. Boyadzhiev, I. V. Feda, Dmitry Fleischman, Robert Frontcza, G. C. Greubel, Albert Natian, Hieyui Ohtsua, Ángel Plaza, Raphael Schumacher (student), Jaroslav Seibert, Jason L. Smith, Albert Stadler, Santiago Alzate Suárez (student), and the proposer. An Inequality with a Geometric Twist B-3 Proposed by Ivan V. Feda, Vasyl Stefany Precarpathian National University, Ivano-Franivs, Uraine. 84 VOLUME 57, NUMBER

5 For all positive integers n and a, prove that n F (F+ a + F + a F n+ a ) 0. ELEMENTARY PROBLEMS AND SOLUTIONS Solution by Wei-Kai Lai, University of South Carolina Salehatchie, Walterboro, SC. The claimed inequality is equivalent to n F (F+ a + F + a ) (F n+ a + ) n F (Fn+ a + )(F n+ ). We find n F (F+ a + F + a ) F F a + F n Fn+ a + (F + F + )F+ a + F nfn+ a + F a+ +. Since F F, we can further rewrite the claimed inequality as or n+ F n Fn+ a + Fi a+ Fn+ a+ F n+ a + F n+, i i n+ Fi a+ Fn+F a n+ Fn+ a + F n+. We will prove this inequality by induction on n. The equality becomes an equality when n. Assume it is true when n. Then, + Fi a+ F+ a F + F+ a + F + + F a+ +. i To complete the inductive step, it suffices to prove that or equivalently, F a + F + F a + + F + + F a+ + F a +3 F + F a +3 + F +3, F + (F a + ) (F + )(F a +3 F a + ). After factoring F+ a and F +3 a F + a and canceling common factors, the inequality above reduces to a a F a j + F a j +3 F j +, j0 j0 which is obviously true. Therefore, the claimed inequality is true for any positive integer n. Solution by the proposer. The inequality becomes an equality when n, so we shall assume n >. Using F, and the identities F F + F + and n F F n+, we can write the given inequality as n (F + F + ) F + a + F + a (F n+ F ) F n+ a + F a. FEBRUARY 09 85

6 THE FIBONACCI QUARTERLY Let A denote the point ( F, F a ) on the graph of the function f(x) x a, and B denote the point (F, 0). The left side is the sum of the areas of the trapezoids A + A + B + B + from to n. The right side of the inequality above is the area of the trapezoid A A n+ B n+ B. Because f(x) a x is a convex function, it is obvious that the left side is less than or equal to the right side. Also solved by Dmitry Fleischman, and Ángel Plaza. An Intriguing Binomial Sum B-4 Proposed by Hideyui Ohtsua, Saitama, Japan. For any positive integer n, prove that n ( ) n F n F, and n ( ) n L n L. Solution by Kambiz Moghaddamfar (student), Sharif University of Technology, Tehran, Iran. Given G 0 and G, the generalized Fibonacci sequence G 0, G, G,..., is defined recursively by G n G + G n for n. First, we claim that ( n n ) G G n ( ) G i n 0 +. i i To prove this, we apply the well-nown identity [] that ( ) G G i i to obtain n G i0 i [ n ( ] )G i i i0 ( n ) [ n n ( ) ] G 0 + G i i i i ( n ) n G i n i ( ) G 0 + i i + j i i + j i i j0 ( n ) n G i n i ( ) G 0 + i + j. i i i j0 Applying the Hocey-Stic Theorem, we find n i ( ) i + j i j0 ( ) n, i 86 VOLUME 57, NUMBER

7 ELEMENTARY PROBLEMS AND SOLUTIONS from which the claim follows. The proof is completed by substituting in G n F n and G n L n, respectively. Solution by Khristo N. Boyadzhiev, Ohio Northern University, Ada, OH. Let a i, i,, 3,..., be a sequence, and let n ( ) n b n a. The following result was proved in [, 3]: n ( ) n a 0 n b. Using the Binet s formula and the binomial theorem, it is easy to show that n ( ) n n ( ) n F F n, and L L n. Since F 0 0, we immediately obtain n ( ) n F n 0 F. The second identity follows in a similar manner, because L 0, and n ( n ) L L n. References [] S. Vajda, Fibonacci and Lucas Numbers and the Golden Ratio, Dover, 008. [] K. N. Boyadzhiev, Binomial transform and the bacward difference, Advan. Appl. Discrete Math., 3 (04), [3] A. N. t Woord, Solution II to Problem 0490, Amer. Math. Monthly, 06 (999), 588. Also solved by I. V. Feda, Dmitry Fleischman, G. C. Greubel, Albert Natian, Ángel Plaza, Raphael Schumacher (student), Jason L. Smith, Albert Stadler, and the proposer. A Sequence of Matrices with Special Properties B-5 Proposed by Jathan Austin, Salisbury University, Salisbury, MD. Construct a sequence {M n } n of 3 3 matrices with positive entries that satisfy the following conditions: (A) M n is the product of nonzero Fibonacci numbers. (B) The determinant of any submatrix of M n is a Fibonacci number or the product of nonzero Fibonacci numbers. (C) lim n M n+ / M n + α. FEBRUARY 09 87

8 THE FIBONACCI QUARTERLY Solution by Ehren Metcalfe, Barrie, Ontario, Canada. Define a sequence of 3 3 matrices {M n } n such that M n F n+ F n+ F n+ F n+ F n+ F n+. F n+ F n+ F n+ Then, each entry of M n is positive, and M n Fn+ 3 Fn+F n+ F n+ Fn+ + Fn+ 3 (Fn+ Fn+)(F n+ F n+ ) (F n+ F n+ ) (F n+ + F n+ ) F nf n+3 is a product of nonzero Fibonacci numbers. For the lower left submatrix, F n+ F n+ F n+ F n+ F n+ F n+ F n+ (F n+ F n+ )F n+ F n F n+. For the upper right submatrix, F n+ F n+ F n+ F n+ F n+ Fn+ (F n+ F n+ )(F n+ + F n+ ) F n F n+3. Since the upper left and lower right can be obtained from the the submatrix above by interchanging their rows, their determinants differ by a factor of F. All these determinants are products of nonzero Fibonacci numbers. Finally, ( ) M n+ Fn+ lim lim Fn+4 α α ( + α) α + α, n M n n F n+3 as desired. Editor s Note: There are other possible answers. Feda gave proposer presented F n F n+ F n+ F n+3 Fn+ Fn+ Fn+3 as solutions. Also solved by I. V. Feda, and the proposer. F n+4 F n F n F n+ F n F n+ F n F n F n+3, and the 88 VOLUME 57, NUMBER