Download MAS110 Problems for Chapter 2: Summation and Induction

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.
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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)
.
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(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.
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