MATH2069_DiscreteMathematicsAndGraphTheory_2014 Semester 1_tute04s

The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial4(Week5)

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

(a)n3+5n is a multiple of3(i.e.n3+5n=3ℓfor some integerℓ).

Solution:The n=0case holds because03+0=0is a multiple of3(it

is3×0).Suppose that n≥1and that the result is known for n−1,i.e.

(n−1)3+5(n−1)=3ℓ,for some integerℓ.

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 of3,establishing the inductive

step and completing the proof.

(b)5n−4n−1is a multiple of16.

Solution:The n=0case holds because50−4×0−1=0is a multiple

of16.Suppose that n≥1and that the result is known for n−1,i.e.

5n−1−4(n−1)−1=16ℓ,for some integerℓ.

This equation can be rewritten as

5n−1=4n−3+16ℓ.

So

5n−4n−1=5(4n−3+16ℓ)−4n−1=16(n−1+5ℓ), which is a multiple of16,establishing the inductive step and completing the

proof.

e the characteristic polynomial to solve the following recurrence relations:

(a)a n=5a n−1−6a n−2for n≥2,where a0=2,a1=5.

Solution:The characteristic polynomial is x2−5x+6=(x−2)(x−3)

with roots2and3,so the general solution is a n=C12n+C23n for some

constants C1,C2.In our case we have

2=a0=C1+C2and5=a1=2C1+3C2.

Solving yields C1=C2=1,so the solution is

a n=2n+3n.

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

Solution:The characteristic polynomial is x2−4x+3=(x−1)(x−3) with roots1and3,so the general solution is a n=C11n+C23n for some constants C1,C2.In our case we have

−1=a0=C1+C2and2=a1=C1+3C2.

Solving yields C1=−5/2and C2=3/2,so the solution is

3n+1−5

a n=

*(f)a n =6a n −1−12a n −2+8a n −3for n ≥3,where a 0=2,a 1=4,a 2=16.

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

with repeated root 2,so the general solution is

a n =C 12n +C 2n 2n +C 3n 22n

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

2=a 0=C 1,4=a 1=2C 1+2C 2+2C 3,16=a 2=4C 1+8C 2+16C 3,

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

a n =2n (2−n +n 2).

panies 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 B retains 60%of its custom and loses to A the remaining 40%.Let a n denote the market share of company A after n years (thus,that of company B is 1−a n ).

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

Solution:For n ≥1,we have

a n =7

10(1−a n −1)=3

10.

(b)Solve the recurrence relation,in the sense of giving a closed formula for a n ,

in terms of a 0.

Solution:Unravelling the recurrence relation,we get:

a n =

310=310a n −2+410=

310410= 310a n −3+410410= 310 2410

410...= 310

310+1 = 3

10

3

37 37.

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

progression.