D> equipment to homework #6, stillproud.org6A Winter "99Solution come homework#6,stillproud.org6A, Winter, 1999-- by Zhao,HongTextbook:Title: "Discrete Mathematstillproud.org and also Its Applications" third EditionAuthor: Kenneth H. RosenP78:3. What room the terms a0, a1, a2,and a3 the the succession an, whereby an equalsa) 2n + 1?Answer: a0 = 2, a1= 3, a2 = 5, a3 = 9b) (n+1)n+1?Answer: a0 = 1, a1= 4, a2 = 27, a3 = 256c) floor(n/2)?Answer: a0 = 0, a1= 0, a2 = 1, a3 = 1d) floor(n/2) + ceiling(n/2)Answer: a0 = 0, a1= 1, a2 = 2, a3 = 37. What is the value of every of the complying with sums of terms of ageometric progression?a) Sum(3 * 2j), where j = 0, 1, 2, ..., 8Answer: ( 3 * 28+1 -3 ) / ( 2 - 1 )= 1533b) Sum(2j), whereby j = 1, 2, ..., 8Answer: ( 28+1 -1 ) / ( 2 - 1 ) - 20= 510c) Sum( (-3)j ), where j = 2, ..., 8Answer: ( (-3)8+1 -1 ) / ( -3 - 1 )- (-3)0 - (-3)1 = 4923d) Sum( 2 * (-3)j ), wherein j = 0, 1, 2, ..., 8Answer: ( 2 * (-3)8+1 -2 ) / ( -3 -1 ) = 984210. Compute every of the following twin sums.a) SumOveri( SumOverj ( ns - j ) ), whereby i = 1, 2, 3, and also j = 1,2.Answer: (1-1) + (1-2) + (2-1) + (2-2) + (3-1) + (3-2)= 3b) SumOveri( SumOverj ( 3i + 2j ) ), wherein i = 0, 1, 2, 3, and j= 0, 1, 2.Answer: (3*0+2*0) + (3*0+2*1) + (3*0+2*2) + (3*1+2*0)+ (3*1+2*1) + (3*1+2*2) + (3*2+2*0) + (3*2+2*1) + (3*2+2*2) + (3*3+2*0)+ (3*3+2*1) + (3*3+2*2) = 78c) SumOveri( SumOverj ( j ) ), wherein i = 1, 2, 3, and j = 0, 1, 2.Answer: 3 * ( 0 + 1 + 2 ) = 9d) SumOveri( SumOverj ( i2j3 ) ), whereby i= 0, 1, 2, and also j = 0, 1, 2, 3.Answer: (02 * 03)+ (02 * 13) + (02 *23) + (02 * 33) +(12 * 03) + (12 *13) + (12 * 23) +(12 * 33) + (22 *03) + (22 * 13) +(22 * 23) + (22 *33) = 180A-2:2. Find each of the adhering to quantities.a) log2 1024Answer: 10b) log2 (1/4)Answer: -2c) log4 8Answer: (log2 8) / (log2 4) = 3/24. Permit a, b, and c be hopeful real numbers. Show that a^(logbc)= c^(logba). Whereby "^" way "power".Proof: logb c = loga c / logab => a^(logb c) = a^(loga c / loga b)= (a^loga c)^(1/loga b) = c^(1/logab) = c^(logb a)where, 1/loga b = 1/(logx b / logxa) = logx a / logx b = logb aP209:3. Find f(2), f(3), f(4), and f(5) if f is identified recursively byf(0) = -1, f(1) = 2 and also for n = 1, 2, ...a) f(n+1) = f(n) + 3 * f(n-1)Answer:f(0) = -1f(1) = 2f(2) = f(1) + 3 * f(0) = 2 + 3 * (-1) = -1f(3) = f(2) + 3 * f(1) = -1 + 3 * 2 = 5f(4) = f(3) + 3 * f(2) = 5 + 3 * (-1) = 2f(5) = f(4) + 3 * f(3) = 2 + 3 * 5 = 17b) f(n+1) = f(n)2 * f(n-1)Answer:f(0) = -1f(1) = 2f(2) = f(1)2 * f(0) = 22 * (-1) = -4f(3) = f(2)2 * f(1) = (-4)2 * 2 = 32f(4) = f(3)2 * f(2) = 322 * (-4) =-4096f(5) = f(4)2 * f(3) = (-4096)2 * 32= 536870912c) f(n+1) = 3 * f(n)2 - 4 * f(n-1)2Answer:f(0) = -1f(1) = 2f(2) = 3 * f(1)2 - 4 * f(0)2 = 3 *22 - 4 * (-1)2 = 8f(3) = 3 * f(2)2 - 4 * f(1)2 = 3 *82 - 4 * 22 = 176f(4) = 3 * f(3)2 - 4 * f(2)2 = 3 *1762 - 4 * 82 = 92672f(5) = 3 * f(4)2 - 4 * f(3)2 = 3 *926722 - 4 * 1762 = 25764174848d) f(n+1) = f(n-1) / f(n)Answer:f(0) = -1f(1) = 2f(2) = f(0) / f(1) = (-1) / 2 = -1/2f(3) = f(1) / f(2) = 2 / (-1/2) = -4f(4) = f(2) / f(3) = (-1/2) / (-4) = 1/8f(5) = f(3) / f(4) = (-4) / (1/8) = -326. Offer a recursive meaning of the succession an, n =1, 2, 3, ... If(Note: There room many feasible correctanswers.)a) an = 4n-2Answer:an-1 = 4(n-1) - 2 = 4n - 6an = 4n - 2 = ( 4n - 6 ) + 4 = an-1+ 4, because that n>=1 and also a1 = 2b) an = 1 + (-1)nAnswer:an-1 = 1 + (-1)n-1an = 1 + (-1)n = 1 + (-1)(n-1)+1= 1 + (-1)n-1 * (-1) = 1 - (-1)n-1 =1 - < 1 + (-1)n-1 - 1 > = 1 - ( an-1 - 1 ) = 2-an-1, for n>=1 and also a1 = 0c) one = n(n+1)Answer:an-1 = (n-1)n = n2 - nan = n(n+1) = n2 + n = (n2- n) + 2n = an-1 + 2n, for n>=1 and a1= 2d) one = n2Answer:an-1 = (n-1)2 = n2- 2n + 1an = n2 = ( n2- 2n + 1 ) + 2n - 1 = an-1 + 2n - 1, because that n>=1 and also a1= 111. Prove the f1 + f3 + ... + f2n-1= f2n at any time n is a hopeful integer. ( fn is thenth Fibonacci number.)Proof: Prove by induction. Permit P(n) it is in " f1+ f3 + ... + f2n-1 = f2n", where n = 1, 2, 3, ...Basis step: f1 = f2 = 1 => P(1) istrue.Inductive step: i think P(n) is true, i.e. F1 + f3+ ... + f2n-1 = f2nThen f1 + f3 + ... + f2n-1+ f2n+1= f2n + f2n+1= f2n+2= f2(n+1)The critical equation mirrors that P(n+1) is true. This completes the inductivestep and also completes the proof.20. Display that the set S characterized by "1 belongs to S" and "s+t belongsto S" anytime "s is an element of S" and "t is an facet of S" is theset of hopeful integers.Proof: permit A be the collection of all hopeful integers. Toprove the A = S, us must display that A is a subset of S and that S is asubset of A.Case 1: A is a subset that S.To prove that A is a subset the S, us must display that every positiveinteger is in S. Us will use mathematical induction come prove this.Let P(n) it is in the statement the " every hopeful integer n belonging toS "Basis step: 1 belongs come S. => P(1) is true, because it is the very first partof the recusive an interpretation of S.Inductive step: assume P(n) is true, i.e. N is in S. Because n is in S and1 is in S, it complies with from the second part of the recursive definitionof S the n+1 is additionally in S. This method P(n+1) is true. For this reason A is a subsetof S.Case 2: S is a subset the A.To prove the S is a subset that A, we usage the recursive meaning ofS. First the basis step of the definition specifies that 1 is in S. Since1 is a hopeful integer, all facets specified to it is in in S in this stepare in A. To complete the proof, we must show that every integers in S generatedusing the second part of the recursive an interpretation are in A. This consistsof reflecting that s+t is in A at any time s and also t are aspects of S also assumedto be in A. Now if s and t room both in A, it adheres to that both s and also tare optimistic integers. Clear s+t is additionally a hopeful integer, for this reason wecompleted the proof.25.


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Specify well-formed fomulae that sets, variables representing sets,and operator from -, U, ^, -, wherein "-" means"complement", "U" method "Union", "^" means "intersection".Answer: If x is a set or a variable representiing a set,then x is a well-formed formula. If x and also y are well-formed formulae, thenso room x-, (x U y), (x ^ y), and also (x - y), wherein "x-"means"complement that x".Office Hours: Monday 3:30pm -- 5:30pm