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Running head: PROBLEMS ON DISCRETE MATHEMATICSPROBLEMS ON DISCRETE MATHEMATICSName of the StudentName of the UniversityAuthor Note

1PROBLEMS ON DISCRETE MATHEMATICS1. (a)The given difference equation is,un=32nun−1+3(2n+1−1)and the initial value,u0=1.Let us assume thatun=d∀nbelonging to the set of Natural numbers.Then, d =32nd+3(2n+1−1).Or, d =3(2n+1−1)1−32nFor the homogeneous part,un−32nun−1=0.Its solution isun=C¿, for some C belonging to the set of Real numbers.Thus the general solution is given by (Atici and Eloe 2015)un=9n2C+3(2n+1−1)1−32n...(1)Since,u0=1,1 =C+9−11−3Or,1 =C+8−2Or, 1 =C−4Thus,C=5.Substituting the value of C in (1), the general solution can be obtained asun=5(9n2)+3(2n+1−1)1−32n.

2PROBLEMS ON DISCRETE MATHEMATICS1. (b)The given difference equation is,un=6un−1−5un−2−32n+44and the initial values are ,u0=9,u1=25.For the homogeneous part,un=6un−1−5un−2Considerun=Awn.In that case,Awn=6Awn−1−5Awn−2Or, 1 =6w−5w2Or,w2=6w−5Or,w2−6w−5=0Or,(w−3)(w−1)=0Thus,w=5∨1.Therefore, solution of the homogeneous part isA1(w1)n+A2(w2)n=A1(1)n+A2(5)n=A1+5nA2Now we consider the general solution asun=A1+5nA2+f(n)Or,un=A1+5nA2+a+bn+cn2; wheref(n)=¿a+bn+cn2Now substitutingf(n)in the given equation,a+bn+cn2=6[a+b(n−1)+c(n−1)2]- 5[a+b(n−2)+c(n−2)2] -32n+44Or,a+bn+cn2=6a+6b+6c−5(a−2b+4c)+[b(a−2c)−5(b−4c)]+n2c−32n+44By comparing we get,a=0+4b−14c+44

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