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The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 3 (Week 4)

MATH2069/2969: Discrete Mathematics and Graph Theory Semester 1, 2012

1. Prove by induction that, for all n ≥ 0,

(a) n3 + 5n is a multiple of 3 (i.e. n3 + 5n = 3ℓ for some integer ℓ).

Solution: The n = 0 case holds because 03 + 0 = 0 is a multiple of 3 (itis 3× 0). Suppose that n ≥ 1 and that the result is known for n− 1, i.e.

(n− 1)3 + 5(n− 1 ) = 3ℓ, for some integer ℓ.

Then

3ℓ = n3 − 3n2 + 3n− 1 + 5n− 5 = n3 + 5n− 3(n2 − n + 2) ,

so n3 + 5n = 3(ℓ + n2 − n + 2) is a multiple of 3, establishing the inductivestep and completing the proof.

(b) 5n − 4n− 1 is a multiple of 16.

Solution: The n = 0 case holds because 50 − 4 × 0 − 1 = 0 is a multipleof 16. Suppose that n ≥ 1 and that the result is known for n− 1, i.e.

5n−1 − 4(n− 1)− 1 = 1 6ℓ, for some integer ℓ.

This equation can be rewritten as

5n−1 = 4n− 3 + 16ℓ.

So5n − 4n− 1 = 5(4n− 3 + 16ℓ)− 4n− 1 = 16(n− 1 + 5ℓ) ,

which is a multiple of 16, establishing the inductive step and completing theproof.

2. Use the characteristic polynomial to solve the following recurrence relations:(a) an = 5an−1 − 6an−2 for n ≥ 2, where a0 = 2, a1 = 5.

Solution: The characteristic polynomial is x2 − 5x + 6 = (x − 2)(x − 3)with roots 2 and 3, so the general solution is an = C 12n + C 23n for someconstants C 1, C 2. In our case we have

2 = a0 = C 1 + C 2 and 5 = a1 = 2C 1 + 3C 2.

Solving yields C 1 = C 2 = 1, so the solution is

an = 2n

+ 3n

.

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(b) an = 4an−1 − 3an−2 for n ≥ 2, where a0 = −1, a1 = 2.

Solution: The characteristic polynomial is x2 − 4x + 3 = (x − 1)(x − 3)with roots 1 and 3, so the general solution is an = C 11n + C 23n for someconstants C 1, C 2. In our case we have

−1 = a0 = C 1 + C 2 and 2 = a1 = C 1 + 3C 2.

Solving yields C 1 = −5/2 and C 2 = 3/2, so the solution is

an = 3n+1 − 5

2 .

(c) an = 4an−1 − 4an−2 for n ≥ 2, where a0 = 3, a1 = 8.

Solution: The characteristic polynomial is x2 − 4x + 4 = (x − 2)2 withrepeated root 2, so the general solution is an = C 12n + C 2n2n for some

constants C 1, C 2. In our case we have

3 = a0 = C 1 and 8 = a1 = 2C 1 + 2C 2,

yielding C 1 = 3 and C 2 = 1, so the final solution is

an = 3× 2n + n2n = (n + 3)2n.

(d) an = 6an−1 − 9an−2 for n ≥ 2, where a0 = 2, a1 = −3.

Solution: The characteristic polynomial is x2 − 6x + 9 = (x − 3)2 withrepeated root 3, so the general solution is an = C 13n + C 2n3n for someconstants C 1, C 2. In our case we have

2 = a0 = C 1 and − 3 = a1 = 3C 1 + 3C 2,

yielding C 1 = 2 and C 2 = −3, so that the final solution is

an = 3n(2 − 3n).

*(e) an = 6an−1 − 11an−2 + 6an−3 for n ≥ 3, where a0 = 3, a1 = 5, a2 = 11.

Solution: The characteristic polynomial is x3−6x2+11x−6 = (x−1)(x−

2)(x− 3) with roots 1, 2, 3, so the general solution is

an = C 1 + C 22n + C 33n

for some constants C 1, C 2, C 3. In our case we have

3 = a0 = C 1 + C 2 + C 3, 5 = a1 = C 1 + 2C 2 + 3C 3, 11 = C 1 + 4C 2 + 9C 3,

yielding C 1 = 2, C 2 = 0 and C 3 = 1, so the final solution is

an = 3n + 2.

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*(f) an = 6an−1 − 12an−2 + 8an−3 for n ≥ 3, where a0 = 2, a1 = 4, a2 = 16.

Solution: The characteristic polynomial is x3 − 6x2 + 12x− 8 = (x− 2)3

with repeated root 2, so the general solution is

an = C 12n + C 2n2n + C 3n22n

for some constants C 1, C 2, C 3. In our case we have

2 = a0 = C 1, 4 = a1 = 2C 1 + 2C 2 + 2C 3, 16 = a2 = 4C 1 + 8C 2 + 16C 3,

yielding C 1 = 2, C 2 = −1 and C 3 = 1, so the final solution is

an = 2n(2− n + n2).

3. Companies A and B control the market for a certain product. From one year to

the next, A retains 70% of its custom and loses to B the remaining 30%, while Bretains 60% of its custom and loses to A the remaining 40%. Let an denote themarket share of company A after n years (thus, that of company B is 1−an).

(a) Write down a recurrence relation expressing an in terms of an−1, for n ≥ 1.

Solution: For n ≥ 1, we have

an = 7

10an−1 +

4

10(1 − an−1) =

3

10an−1 +

4

10.

(b) Solve the recurrence relation, in the sense of giving a closed formula for an,in terms of a0.

Solution: Unravelling the recurrence relation, we get:

an = 3

10an−1 +

4

10 =

3

10

3

10an−2 +

4

10

+

4

10

=

3

10

2

an−2 + 3

10

4

10 +

4

10 =

3

10

2 3

10an−3 +

4

10

+

3

10

4

10 +

4

10

=

3

10

3

an−3 +

3

10

2 4

10 +

3

10

4

10 +

4

10...

=

3

10

n

a0 + 4

10

3

10

n−1

+ · · ·+ 3

10 + 1

=

3

10

n

a0 + 4

10

3

10

n− 1

3

10 − 1

=

a0 −

4

7

3

10

n

+ 4

7.

Here the second-last equality uses the formula for the sum of a geometric

progression.

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(c) Hence prove that the market share of company A in the long run (i.e. thelimit of an as n→∞) is independent of its initial market share a0.

Solution: As n →∞, the power ( 310

)n tends to 0, so an → 4

7. So whatever

the initial situation, the market tends to a stable situation where company

A has a 4

7 market share and company B has a 3

7 market share.

4. Let bn be the number of ways of forming a line of n people distinguished only bywhether they are male (M) or female (F), such that no two males are next to eachother. For example, the possibilities with 3 people are FFF, FFM, FMF, MFF,and MFM, so b3 = 5. Write down a recurrence relation for bn. Do you recognizethe sequence?

Solution: We have b0 = 1, b1 = 2, and bn = bn−1 + bn−2 if n ≥ 2. To seethis notice that in a line of n people with n ≥ 2, if the last person is female thenthere are bn−1 possibilities for the line of the first n − 1 people, whilst if the last

person is male then the second last person must be female, so that there are bn−2possibilities for the line of the first n − 2 people. We get the Fibonacci sequencewith the first two terms deleted, so bn = F n+2.

5. Define a sequence recursively by a0 = 1, a1 = 2, and an = an−1an−2 for n ≥ 2.

(a) Find a2, a3, a4, a5 and a6.

Solution: a2 = 2, a3 = 4 = 22, a4 = 8 = 23, a5 = 32 = 25, a6 = 28.

(b) Prove that an = 2F n, where F 0, F 1, F 2, · · · is the Fibonacci sequence.

Solution: As seen in lectures, we only need to show that 2

F n

satisfies thesame initial conditions and recurrence relation as an. The initial conditionshold because 2F 0 = 20 = 1 and 2F 1 = 21 = 2. The recurrence relation holdsbecause for n ≥ 2,

2F n = 2F n−1+F n−2 = 2F n−12F n−2,

by the Fibonacci recurrence relation F n = F n−1 + F n−2.

6. Imagine a 2n×2n array of equal-sized squares, where n is some positive integer. Wewant to cover this array with non-overlapping L-shaped tiles, each of which exactly

covers three squares (one square and two of the adjacent squares, not opposite toeach other). Since the number of squares is not a multiple of 3, we need to removeone square before we start. Prove by induction that no matter which square weremove, the remaining squares can be covered by these L-shaped tiles.

Solution: The base case is clear, because removing a square from a 2× 2 arrayleaves 3 squares which can be covered by a single tile. We now prove the claimfor n ≥ 2, assuming its truth for n − 1. Let G be the 2n × 2n array with exactlyone square missing. Denote the quarters of G by UL for upper left, UR for upperright, LL for lower left and LR for lower right. Each quarter is a 2n−1×2n−1 array,

except that one of the quarters has one square missing. By rotating G if necessary,we may suppose that the missing square is in LL. Let T be the L-shape formed bythe lower-rightmost subsquare of UL, the lower-leftmost subsquare of UR and the

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upper-leftmost subsquare of LR. Then removing T from G produces a union of four 2n−1×2n−1 arrays each with one square missing, and each of these can be tiledby L-shapes, by the induction hypothesis. Hence G is tiled by all these L-shapestogether with T , establishing the inductive step and completing the proof.

*7. The following argument ‘proves’ that whenever a group of people is in the sameroom, they all have the same height. There must be an invalid step; find it.

We argue by induction on the number n of people in the room. Then = 1 case is obviously true. Suppose that n ≥ 2 and that the claimholds for rooms with n − 1 people. Let P 1, P 2, . . . , P n be the n peoplein this room. If P n were to leave the room we would have a room withn−1 people, so by the inductive hypothesis, P 1, P 2, . . . , P n−1 all have thesame height. We can apply the same reasoning with P 1 leaving the room,so P 2, . . . , P n−1, P n all have the same height. But P 2 is in both these

collections, so all of P 1, P 2, . . . , P n have the same height. This establishesthe inductive step, and so the claim holds for all n by induction.

Solution: Since the claim is false even when n = 2, the proof must fail alreadyin this case; when you run through the argument with n = 2, the error emerges.The invalid step is the assertion that “P 2 is in both these collections”, because thisignores the convention governing the way these collections were written out. Whenyou start with n people P 1, P 2, . . . , P n and remove P n, it is reasonable to list theremaining people as “P 1, P 2, . . . , P n−1”, but you have to bear in mind that if n = 2,this list will just consist of P 1 and will not in fact include P 2.

*8. For which n is the Fibonacci number F n even, and for which n is F n odd? Proveyour answer by induction.

Solution: Examining the first few terms, one is led to guess that F n is evenwhen n is a multiple of 3, and odd when n is not a multiple of 3. To prove thisby induction, we first observe that the n = 0 and n = 1 cases are true (becauseF 0 = 0 is even and F 1 = 1 is odd). Then in proving the result for n ≥ 2, we canassume it for n − 1 and for n− 2. Recall that we have

F n = F n−1 + F n−2.

There are now three cases, depending on the remainder of n after division by 3.

If n ≡ 0 (mod 3) (i.e. n is a multiple of 3), then n − 1 and n − 2 are not multiplesof 3, so F n−1 and F n−2 are odd, so F n is even as required.

If n ≡ 1 (mod 3), then n − 1 is a multiple of 3 but n − 2 is not, so F n−1 is evenand F n−2 is odd, so F n is odd as required.

If n ≡ 2 (mod 3) then n − 1 is not a multiple of 3 but n − 2 is, so F n−1 is odd andF n−2 is even, so F n is odd as required.

This completes the inductive step, and the claim follows by induction.

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**9. Suppose we want to solve a recurrence relation which is almost a kth-order homo-geneous linear recurrence relation, but with an extra constant term C :

an = r1an−1 + r2an−2 + · · ·+ rkan−k + C, for all n ≥ k.

Let p(x) = xk

−r1xk−1

−···−rk be the characteristic polynomial of the homogeneousrecurrence relation obtained by omitting C .

(a) Show that any solution an also satisfies the (k + 1)th-order linear homoge-neous recurrence relation with characteristic polynomial (x− 1) p(x).

Solution: Suppose that the sequence an is a solution of our recurrencerelation. Then for any n ≥ k + 1, we have

an = r1an−1 + r2an−2 + · · ·+ rkan−k + C and

an−1 = r1an−2 + r2an−3 + · · ·+ rkan−k−1 + C.

Subtracting the second equation from the first gives

an − an−1 = r1(an−1 − an−2) + r2(an−2 − an−3) + · · ·+ rk(an−k − an−k−1)

for all n ≥ k + 1, which can be rearranged as

an = (r1 − 1)an−1 + (r2 − r1)an−2 + · · ·+ (rk − rk−1)an−k + (−rk)an−k−1.

This is the homogeneous recurrence relation with characteristic polynomial

xk+1 − (r1 − 1)xk − (r2 − r1)xk−1 − · · · − (rk − rk−1)x + rk

= (x− 1)(xk − r1xk−1 − · · · − rk−1x− rk) = (x− 1) p(x),

as claimed.

(b) Hence describe the general solution an in terms of the roots of p(x). (Theanswer will depend on whether 1 is a root of p(x) or not.)

Solution: Let the different roots of p(x) be λ1, · · · , λs with multiplicitiesm1, · · · , ms (where the multiplicity of a non-repeated root is 1).

First suppose that none of the λi’s equals 1; then (x − 1) p(x) has rootsλ1, · · · , λs, 1 with multiplicities m1, · · · , ms, 1. By the general solution of homogeneous recurrence relations given in lectures, (a) implies that

an = (C 11 + C 12n + · · ·+ C 1,m1nm1

−1)λn1 + · · ·

+ (C s1 + C s2n + · · ·+ C s,msnms−1)λns + D,

for some constants C ij, D. Conversely, any sequence an of this form is a solu-tion of the homogeneous recurrence relation with characteristic polynomial(x−1) p(x). This is not quite enough to imply that it satisfies our recurrencerelation: but the additional requirement is just the n = k case, namely

ak = r1ak−1 + · · ·+ rk−1a1 + rka0 + C,

because, as we saw in the previous part, the difference between this equationand the n = k + 1 case is a case of the homogenous recurrence relation, as isthe difference between the n = k + 1 case and the n = k + 2 case, and so on.

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It is easy to see that the n = k case reduces to a constraint on the constantD, namely

D = r1D + · · ·+ rkD + C,

so the general solution is given by the above formula but with D specified

to equal C

1− r1 − · · · − rk . (The denominator is p(1), which we assumed to

be nonzero.)

Now suppose that 1 is a root of p(x); without loss of generality, λs = 1. Then(x−1) p(x) has roots λ1, · · · , λs−1, 1 with multiplicities m1, · · · , ms−1, ms+1.Solving the homogeneous recurrence, we obtain

an = (C 11 + C 12n + · · ·+ C 1,m1nm1−1)λn1 + · · ·

+ (C s−1,1 + C s−1,2n + · · ·+ C s−1,ms−1−1nms−1−1)λns−1

+ (C s1 + C s2n + · · ·+ C s,ms−1nms−1 + Dnms),

for some constants C ij, D. As in the previous case, we have one extra con-straint on the constant D in order that the n = k case of the desired recur-rence relation should hold, namely

Dkms = r1D(k − 1)ms + · · ·+ rk−1D + C,

so the general solution is given by the above formula but with D specified to

equal C

kms − r1(k − 1)ms − · · · − rk−1. (The denominator is nonzero, because

it is what you get when you substitute x = 1 in the polynomial obtainedfrom p(x) by applying ms times the operator x d

dx

; each application reducesthe multiplicity of the root 1 by 1, so it is no longer a root at the end.)

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