ADVANCED PROBLEMS AND SOLUTIONS PROBLEMS PROPOSED IN THIS ISSUE

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1 EDITED BY LORIAN LUCA Please sed all commuicatios cocerig to LORIAN LUCA, SCHOOL O MATHEMATICS, UNIVERSITY O THE WITWA- TERSRAND, PRIVATE BAG X3, WITS 00, JOHANNESBURG, SOUTH ARICA or by at the address florialuca@witsacza as files of the type tex, dvi, ps, doc, html, pdf, etc This departmet especially welcomes problems believed to be ew or extedig old results Proposers should submit solutios or other iformatio that will assist the editor To facilitate their cosideratio, all solutios set by regular mail should be submitted o separate siged sheets withi two moths after publicatio of the problems PROBLEMS PROPOSED IN THIS ISSUE H-87 Proposed by Hideyuki Ohtsuka, Saitama, Japa or, fid closed form expressios for the sums i k k k+ ; ii iii iv k 3L k L k+ L ; k kl k L k+ L ; k G k +kl k L k+ L, where {G } satisfies G + = G + + G for with arbitrary G ad G H-88 Proposed by Hideyuki Ohtsuka, Saitama, Japa Determie ad EBRUARY 08 89

2 THE IBONACCI QUARTERLY H-89 Proposed by D M Bătieţu-Giurgiu, Bucharest, ad Neculai Staciu, Buzău, Romaia Let f : R R be a cotiuous ad odd fuctio ad g : R + R be a cotiuous fuctio such that g/x = gx for all x R + Compute α β where α = + / ad β = / dx + x + e f gx, H-80 Proposed by D M Bătieţu-Giurgiu, Bucharest, ad Neculai Staciu, Buzău, Romaia If a, b, c R +, compute lim + a+ a+ +!!+ b!! b!l c a SOLUTIONS Closed forms for sums of series ivolvig reciprocals of shifted iboacci squares H-783 Proposed by Hideyuki Ohtsuka, Saitama, Japa Vol 4, No, ebruary 06 Prove that i + = 3 + ; 6 ii iii =3 =3 4 = 43 ; 8 = 3 8 Solutio by Ágel Plaza i We will show that =0 + = α = +, ad that =0 + + = 3 These two series are cosequeces of the followig two idetities that may be proved by iductio: m + = m m+, m+ + + = 4m+4/3 m+ m+3 =0 Therefore, the sum proposed i i is + = + + =0 =0 + + = α + 3 = VOLUME 6, NUMBER

3 ii Sice 4 = / / +, the =3 = =3 = = = 43 8 where we have used the sum give i iii, which is proved below iii irst, ote that 4 = + + ad that = + + Therefore, 3 4 = /3 / Takig ito accout the followig relatio equatio 4 i [: it is deduced that i= i i+ i+ i+3 = 7 4 =3 =3 from where the sum iii follows /3 = /3 + + = α, 7 4 α 6 [ R S Melham, iite sums that ivolve reciprocal of products of geeralized iboacci umbers, Itegers, 34 03, A40 Also solved by Bria Bradie, Dmitry leischma, ad the proposer A pair of idetities for π H-784 Proposed by Gleb Glebov, Simo raser Uiversity, Caada Vol 4, No, ebruary 06 Prove that [ i 4k + 4k + 4k + 4k [ ii 4k + 7 4k 7 + 4k + 4k Solutio by Hideyuki Ohtsuka It is kow that πx cot πx =, = π 6 + = π 6 x k x ; 3 EBRUARY 08 9

4 THE IBONACCI QUARTERLY rom the above idetity, we have πx cot πx 4k = 4x 4x i Note that We have ii Note that We have cot π 4 = ad cot π 4 = LHS = = 4k 4k π π cot 4 4 π 4 cot π 4 = + π π = RHS cot 7π 4 = ad cot π 4 = LHS = 4 = 4 7 4k 7 0 4k 7π 7π cot π π cot 4 4 = 7 + π π = RHS Also solved by Bria Bradie, Keeth B Daveport, Dmitry leischma, David Terr, Nicuşor Zlota, ad the proposer 9 VOLUME 6, NUMBER

5 Sums of iboomial coefficiets H-78 Proposed by Hideyuki Ohtsuka, Saitama, Japa Vol 4, No, ebruary 06 Let deote the iboomial coefficiet or m, fid closed forms expressios k for the sums m i k ; + k m ii k + k Solutio by the proposer It is kow that a+r b+r r a b = a+b+r r see [0a Puttig a = s k, b = t k, ad r = k i the above idetity, we have i We have Therefore, we have s+k t+k sk tk = s+t k 3 m m + k + k = +k m+k m k + k m + k = +k m+k k mk m m + k = +m k m by 3 m + k mk m m m k + k = m [ m m +m + k + k = [ m m m +m m m + = m +m m m m m = m +m m m EBRUARY 08 93

6 THE IBONACCI QUARTERLY ii We have + + k + m + + m k = +k+ m+k+ m k+ + + k m+ + + k = +k+ m+k+ +k m+k m + m+ + k = +m+ k m by 3 + m+ + k Therefore, we have Note: k + k = + m+ +m+ = + m+ +m+ = + m+ +m+ = + +m+ +m+ mk+ m m+ m [ + m + + m + + k k [ + m + + m + + m + + m [ m++ m + m+ m m+ m + + m+ m m m + m m + m+ +m+ Similarly, for positive itegers ad r we obtai k r + k r k = r r r [ S Vajda, iboacci ad Lucas umbers ad the golde sectio, Dover, 008 The area of a iboacci polygo H-786 Proposed by Atara Shriki, Oraim College of Educatio Vol 4, No, ebruary 06 Assume that the cosecutive umbers i the iboacci sequece are the coordiates of a polygo s vertices i the Cartesia coordiate system, couterclockwise: A, ; A 3, 4 ; A 3, 6 ; A 4 7, 8 ; ; A, What is the area of such a polygo? 94 VOLUME 6, NUMBER

7 Solutio by Virgiia Johso Oe formula for area bouded by a polygo with coordiates with vertices at P x, y, P x, y,, P x, y is the so called shoelace formula or surveyor s formula, give by the absolute value of x y + x y x y + x y y x y x 3 y x y x See referece [ Takig the vertices i couterclockwise order, the area of the polygos is A = Reorderig the terms, we have A = Note that after the first pair, each of the subsequet pairs have the form j j j j3 Usig a idetity from Everma, et al [: we have that equatio 4 reduces to +k +h +h+k = h k, A = = + Therefore, the area of the polygo is + = + [ B Brade, The surveyor s area formula, The College Mathematics Joural, , [ D Everma, A Daese, K Vekaayah, ad E Scheuer, Elemetary problems ad solutios: Some properties of iboacci umbers, The America Mathematical Mothly, , 694 Also solved by Harris Kwog, Ágel Plaza, ad the proposer Errata: I the statemet of H-8, the coditio p > must be added Withdrawals: Problem H-86 is withdraw as beig a particular case of B-73 EBRUARY 08 9