By Ian Anderson

Discrete arithmetic has now validated its position in so much undergraduate arithmetic classes. This textbook presents a concise, readable and available advent to a few subject matters during this sector, reminiscent of enumeration, graph idea, Latin squares and designs. it really is aimed toward second-year undergraduate arithmetic scholars, and offers them with the various uncomplicated recommendations, principles and effects. It comprises many labored examples, and every bankruptcy ends with numerous workouts, with tricks or suggestions supplied for many of them. in addition to together with general issues similar to binomial coefficients, recurrence, the inclusion-exclusion precept, bushes, Hamiltonian and Eulerian graphs, Latin squares and finite projective planes, the textual content additionally comprises fabric at the ménage challenge, magic squares, Catalan and Stirling numbers, and event schedules.

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**Example text**

But here is another method . 2 ) This aga in is a second ord er recurrence relation; we now show how to solve it. 23 2. 3) where A, B are constants, B i 0, and where al and a2 are given. 3) is called a second order linear recurrence relation with constant coefficients; it turns out that there is a very neat method of solving such recurrences. 3)? e. 0 2 = Aa + B . 3) precisely when a is a solution of the auxiliary equation ° x 2=Ax+B. 3) . If x 2 - Ax - B = (x - a)2 so that A = 2a and B Aan-l = _a 2.

1) for the flags by iteration . Exercise 2. 7 The Lucas numbers L n are defined by L 1 (n 2: 3) . 4, first eliminating 1 and then eliminating powers of 2. You should obtain an - 5a n-l + 8a n-2 - 4a n- 3 = O. 9 Verify tha t if u;, and u ~ are t wo solut ions of the rec urrence an = A an_1 + B a ll - 2 t hen a;, + a;; is also a solution. 10 Show t ha t t he generating fun ction for t he Fibonacci sequence is ti ~~:~ . 7). 11 Let M = (~i) . (a ) Prove that M n+2 = ( r~n,, + l F n0 +' ) wbere Fn is t he n th Fi bona cci num +2 ber.

4) . Then (i) if a i f3, there are constants Ks , K2 such that an = Kso" n 2: 1; (ii) if a = f3, there are constants K 3 , K 4 such that an = (K3 n2:l. e. 5) 24 Discrete Math ematic s Th en t he asserti on th at an = K 1an + K 2 /3 n is certainly true for n = 1, 2. We now pr oceed by ind uction . Assum e th e assertio n is t ru e for all n ::; k . Th en ak+1 = Aa k + Bak _ 1 = A(K1a k + K 2 /3 k) + B (K1a k- 1 + K 2 /3k-l ) = K 1a k- 1 (A a + B ) + K 2 /3 k-I (A/3 + B ) so the result follows. Th en th e assertion that an = (K 3 + n K 4)a n is certainly true for n Assume it is true for all n ::; k.