* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download MAS110 Problems for Chapter 2: Summation and Induction
Survey
Document related concepts
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Infinitesimal wikipedia , lookup
Abuse of notation wikipedia , lookup
Mathematical proof wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Location arithmetic wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Brouwer–Hilbert controversy wikipedia , lookup
Large numbers wikipedia , lookup
Peano axioms wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Collatz conjecture wikipedia , lookup
Transcript
MAS110 Problems for Chapter 2: Summation and Induction 1. (The square pyramid.) Check that the sum of the first 24 square numbers is again a square number. 2. Prove, by induction, that the sum of the first n cubes (of natural numbers) is equal to the square of the sum of the first n natural numbers. 3. What does the sum of the first n odd numbers appear to be? Prove the suspected formula. 4. (i) (ii) (iii) (iv) Use induction to show that n3 − n is divisible by 3 for all n ∈ N. Use induction to show that n5 − n is divisible by 5 for all n ∈ N. Is it true that n4 − n is divisible by 4 for all n ∈ N? (Fermat’s Little Theorem.) Show that if p is a prime number then np − n is divisible by p for all n ∈ N. 5. An arithmetic series with first term 11 and 10 terms has the same sum as a geometric series with first term 5, 10 terms and common ratio 2. What is the common difference d? 6. Re-prove the formula for the sum of a geometric series with first term a, common ratio r 6= 1 and n terms, by induction on n. 7. Show that if a, b, c, d are four consecutive terms of a geometric progression then (b − c)2 = ac + bd − 2ad. 8. I agree to pay £110,000 for a house. After having a survey conducted and seeing how much remedial work is required, I decide to seek a price reduction, and propose a new price of £100,000. The vendor splits the difference and proposes £105,000. Then I split the difference and propose £102,500. If we continue this process of offer and counter-offer, each time splitting the difference, to what limit do our proposed prices converge? (Hint: Use an infinite geometric series.) 9. Prove by induction that dn (xeλx ) = eλx (λn x + nλn−1 ) for all n ∈ N. dxn 10. Use induction to show that the following statements are true for all positive integers n. (i) 2n+1 < 1 + (n + 1)2n . (ii) If x > −1 then (1 + x)n ≥ 1 + nx. 11. Let (Tn )∞ n=1 be the sequence defined by T1 = T2 = T3 = 1 and Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 4. Prove that Tn < 2n for all n ∈ N. 1 12. What is wrong with the following ‘proof’ that any two positive integers are the same? For each n ∈ N = {1, 2, 3, . . .}, let P (n) be the following statement: If a, b ∈ N and n = max{a, b}, then a = b. Base step. If n = 1 and the maximum of two natural numbers a, b is 1 then we must have a = b = 1. So P (1) is true. Inductive step. Assume that P (k) is true for some k ≥ 1. Thus if the maximum of two positive integers is k then those two numbers are the same. Now let a, b be two positive integers such that max{a, b} = k + 1. Then max{a − 1, b − 1} = k. Hence by the inductive hypothesis a − 1 = b − 1, and so a = b. Thus P (k + 1) is true. Conclusion. Hence by induction, P (n) is true for all n ∈ N. 13. Here is one way of figuring out what 14 + 24 + . . . + n4 is in closed form. For integers k ≥ 0 and n ≥ 1, we set Sk (n) := 1k + 2k + . . . + nk . (i) Write down closed expressions for S1 (n), S2 (n) and S3 (n). (ii) By the Binomial Theorem, we have (x + 1)5 − x5 = 5x4 + 10x3 + 10x2 + 5x + 1. Now plug in x = 1, 2, . . . n and add up to obtain (n + 1)5 − 1 = 5S4 (n) + 10S3 (n) + 10S2 (n) + 5S1 (n) + S0 (n). (iii) Substituting in the expressions for S1 (n), S2 (n) and S3 (n), show that 14 + 24 + . . . + n4 = n(n + 1)(2n + 1)(3n2 + 3n − 1) . 30 (iv) When does 12 + 22 + . . . + n2 divide 14 + 24 + . . . + n4 ? 14. The tomatina, sort of. A number of people take part in a tomato throwing contest. They stand in a field in such a way that each person has a unique nearest neighbour, and each person throws a tomato to their nearest neighbour. (This is completely reasonable: distances between contestants will all be different if measured accurately.) Every tomato thrown hits the intended target. The winner or winners, if at all, are those who aren’t hit by a tomato. (i) Give examples to show that if n is even then there might not be a winner. A diagram when n = 6 will do if it is clear how it generalises. (ii) Show that if n ≥ 2 then two contestants will throw tomatoes at each other. Hint: Try a pair of contestants whose distance between them minimises all possible distances between contestants. (iii) Show by induction that if n is odd then there is always at least one winner. Hint: For the inductive step, pick two contestants A and B who throw tomatoes at each other. Then either no other contestant throws a tomato at A or B, or some other contestant throws a tomato at A or B. Consider these cases separately. One case will need the inductive hypothesis; the other needs a direct argument. 2