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Advanced Higher Mathematics – Applications of Algebra and Calculus

Unit Assessment Preparation - Further Practice Questions

1.2 Applying summation formulae

· use mathematical induction to prove summation formulae

1.Use proof by induction to show that, for all ,

(a)(b)(c)

(d) (e)(f)

1.5 Applying algebraic and geometric skills to methods of proof

· Disproving a conjecture by providing a counter-example

2. (a) For any real numbers and , it is conjectured that

and

Find a counter-example to disprove this conjecture.

(b)For any real numbers and it is conjectured that


(c)For any real numbers and it is conjectured that



· Using direct and indirect proof in straightforward examples

3.(a)Prove by contradiction, that if is even then is also even, where

is a natural number.

(b)Prove by contradiction, that if is even then is also even, where


(c)Prove by contradiction, that if is odd then is also odd, where


(d)Prove by contradiction, that if is odd then is also odd, where


(e)Use direct proof to show that if 3 is added to the square of an odd integer

the answer is divisible by 6.

(f)Use direct proof to show that the product of any two odd integer is an even integer.

(g)Use direct proof to show that if 5 is taken away from any odd number the answer is a

multiple of 2.

Answers:

1(a)Assume true for and Consider

LHS =

RHS =

LHS = RHS, so true for

Using , Check for , LHS = , RHS =

LHS = RHS, so true statement for . Hence, if true for implies true for

, and since true for then by induction, statement is true for all .

1(b)Assume true for and Consider

LHS =

RHS = , LHS = RHS, so true for


LHS = RHS, so statement true for . Hence, if true for implies true for

, and since true for then by induction, statement true for all .

1(c)Assume true for and Consider

LHS =





1(d)Assume true for and Consider

LHS =

RHS = LHS = RHS, so true for




1(e)Assume true for and Consider

LHS =





1(f)Assume true for and Consider

LHS =





2(a)Choose suitable values, eg and

Show result is false, eg and

Communicate result, Since conjecture is not true

(b)Choose suitable values, eg , and

Show result is false, eg


(c)Choose suitable values, eg and

Show result is false, and


3 (a)State opening conjecture,

Assume that there is an odd number such that is even

Start process, Let

Complete process, which is odd since

Contradiction statement and conclusion,

This contradicts the initial statement so if is even then is also even.

(b)State opening conjecture,

Assume that there is an odd number such that is even

Start process, Let

Complete process, which is odd since


This contradicts the initial statement so if is even then is also even.

(c)State opening conjecture,

Assume that there is an even number such that is odd

Start process, Let

Complete process, which is even since


This contradicts the initial statement so if is odd then is also odd.

(d)State opening conjecture,

Assume that there is an even number such that is odd

Start process, Let

Complete process, which is even since


This contradicts the initial statement so if is odd then is also odd.

(e)Start process, odd number where

Complete process,

Communicate result, is divisible by 4 since is clearly an integer

(f)Start process, odd integers , where

Complete process,

Communicate result, is odd since is clearly even

(g)Start process, odd number , where

Complete process,

Communicate result, is a multiple of 2

Page 2

6 ×1= 6

6´1=6

3×1(1+1) = 6

3´1(1+1)=6

n = 1

n=1

n = k

n=k

n = k +1

n=k+1

rr=1

n

∑ = n(n +1)2

r

r=1

n

å

=

n(n+1)

2

n = 1

n=1

n∈N

nÎN

8r = 4k(k +1)r=1

k

∑

8r=4k(k+1)

r=1

k

å

8r = 8r + 8(k +1)r=1

k

∑r=1

k+1

∑ = 4k(k +1)+ 8(k +1) = 4(k +1)(k + 2)

8r=8r+8(k+1)

r=1

k

å

r=1

k+1

å

=4k(k+1)+8(k+1)=4(k+1)(k+2)

4(k +1)(k +1+1)⇒ 4(k +1)(k + 2)

4(k+1)(k+1+1)Þ4(k+1)(k+2)

8r = 3n(n +1)r=1

n

∑

8r=3n(n+1)

r=1

n

å

n = 1

n=1

8 ×1= 8

8´1=8

4 ×1(1+1) = 8

4´1(1+1)=8

6rr=1

n

∑ = 3n(n +1)

6r

r=1

n

å

=3n(n+1)

n = 1

n=1

(2r −1) = k2r=1

k

∑

(2r-1)=k

2

r=1

k

å

n = k +1

n=k+1

(2r −1) = (2r −1)+ 2(k +1)r=1

k

∑r=1

k+1

∑ −1= k2 + 2k +1= (k +1)2

(2r-1)=(2r-1)+2(k+1)

r=1

k

å

r=1

k+1

å

-1=k

2

+2k+1=(k+1)

2

(k +1)2

(k+1)

2

n = k +1

n=k+1

(2r −1) = n2r=1

n

∑

(2r-1)=n

2

r=1

n

å

n = 1

n=1

8rr=1

n

∑ = 4n(n +1)

8r

r=1

n

å

=4n(n+1)

2 ×1−1= 1

2´1-1=1

1×1= 1

1´1=1

n = 1

n=1

n = k

n=k

n = k +1

n=k+1

n = 1

n=1

n∈N

nÎN

(6r −1) = k(3k + 2)r=1

k

∑

(6r-1)=k(3k+2)

r=1

k

å

(6r −1) = (6r −1)+ 6(k +1)−1r=1

k

∑r=1

k+1

∑ = k(3k + 2)+ 6(k +1)−1= 3k2 + 2k + 6k + 6 −1= (3k + 5)(k +1)

(6r-1)=(6r-1)+6(k+1)-1

r=1

k

å

r=1

k+1

å

=k(3k+2)+6(k+1)-1=3k

2

+2k+6k+6-1=(3k+5)(k+1)

2r −1= n2r=1

n

∑

2r-1=n

2

r=1

n

å

(k +1)× (3(k +1)+ 2) = (k +1)(3k + 5)

(k+1)´(3(k+1)+2)=(k+1)(3k+5)

(6r −1) = n(3n + 2)r=1

n

∑

(6r-1)=n(3n+2)

r=1

n

å

6 ×1−1= 5

6´1-1=5

1(3×1+ 2) = 5

1(3´1+2)=5

(8r − 5) = k(4k −1)r=1

k

∑

(8r-5)=k(4k-1)

r=1

k

å

(8r − 5) = (8r − 5)+ 8(k +1)− 5r=1

k

∑r=1

k+1

∑ = k(4k −1)+ 8(k +1)− 5 = 4k2 − k + 8k + 8 − 5 = (4k + 3)(k +1)

(8r-5)=(8r-5)+8(k+1)-5

r=1

k

å

r=1

k+1

å

=k(4k-1)+8(k+1)-5=4k

2

-k+8k+8-5=(4k+3)(k+1)

(6r −1)r=1

n

∑ = n(3n + 2)

(6r-1)

r=1

n

å

=n(3n+2)

(k +1)× (4(k +1)−1) = (k +1)(4k + 3)

(k+1)´(4(k+1)-1)=(k+1)(4k+3)

(8r − 5) = n(4n −1)r=1

n

∑

(8r-5)=n(4n-1)

r=1

n

å

8 ×1− 5 = 3

8´1-5=3

1(4 ×1−1) = 3

1(4´1-1)=3

a = −3,

a=-3,

b = 1,

b=1,

c = −5

c=-5

8r − 5 = n(4n −1)r=1

n

∑

8r-5=n(4n-1)

r=1

n

å

d = −2

d=-2

ac = 15

ac=15

bd = −2

bd=-2

ac > bd

ac>bd

a = −3

a=-3

b = −4

b=-4

ab= 34<1

a

b

=

3

4

<1

a > b

a>b

a = −5

a=-5

b = −3

b=-3

a,b,c

a,b,c

a2 = 25

a

2

=25

b2 = 9

b

2

=9

a < b

a

n

n

3n

3n

n = 2k −1,

n=2k-1,

k ∈Ν

kÎN

3n = 3(2k −1) = 6k − 3= 2(3k −1)−1,

3n=3(2k-1)=6k-3=2(3k-1)-1,

3k −1∈Ν

3k-1ÎN

n

n

d

d

3n

3n

5n

5n

5n = 5(2k −1) = 10k − 5 = 2(5k − 2)−1,

5n=5(2k-1)=10k-5=2(5k-2)-1,

5k − 2∈Ν

5k-2ÎN

5n

5n

n2

n

2

n = 2k,

n=2k,

a < b

a

n2 = (2k)2 = 4k2 = 2(2k2 ),

n

2

=(2k)

2

=4k

2

=2(2k

2

),

2k2 ∈Ν

2k

2

ÎN

n

n

2n2

2n

2

n = 2k,

n=2k,

2n2 = 2(2k)2 = 8k2 = 2(4k2 ),

2n

2

=2(2k)

2

=8k

2

=2(4k

2

),

4k2 ∈Ν

4k

2

ÎN

2n2

2n

2

c < d⇒ ac < bd

c

2n −1,

2n-1,

n∈Ν

nÎN

(2n −1)2 = 4n2 − 4n +1+ 3= 4n2 − 4n + 4

(2n-1)

2

=4n

2

-4n+1+3=4n

2

-4n+4

= 4(n2 − n +1)

=4(n

2

-n+1)

n2 − n +1

n

2

-n+1

2n −1

2n-1

(2n −1)(2n −1) = 4n2 − 4n +1= 4(n2 − n)+1

(2n-1)(2n-1)=4n

2

-4n+1=4(n

2

-n)+1

4(n2 − n)+1

4(n

2

-n)+1

4(n2 − n)

4(n

2

-n)

a

a

(2n −1)− 5 = 2n − 6

(2n-1)-5=2n-6

2n − 6 = 2(n − 3)

2n-6=2(n-3)

b

b

ab<1⇒ a < b

a

b

<1Þa

a2 > b2 ⇒ a > b

a

2

>b

2

Þa>b

3n

3n

n

n

n

n

5n

5n

n2

n

2

2n2

2n

2

r = n(n +1)2r=1

n

∑

r=

n(n+1)

2

r=1

n

å

n = k +1

n=k+1

r = r + (k +1)r=1

k

∑r=1

k+1

∑ = k(k +1)2 + (k +1) =k(k +1)+ 2(k +1)

2= (k +1)(k + 2)

2

r=r+(k+1)

r=1

k

å

r=1

k+1

å

=

k(k+1)

2

+(k+1)=

k(k+1)+2(k+1)

2

=

(k+1)(k+2)

2

k(k +1)2

⇒ (k +1)(k +1+1)2

= (k +1)(k + 2)2

k(k+1)

2

Þ

(k+1)(k+1+1)

2

=

(k+1)(k+2)

2

n = k +1

n=k+1

r = n(n +1)2r=1

n

∑

r=

n(n+1)

2

r=1

n

å

n ≥1

n³1

n = 1

n=1

1= 1

1=1

1(1+1)2

= 22= 1

1(1+1)

2

=

2

2

=1

n = 1

n=1

n = k

n=k

n = k +1

n=k+1

n = 1

n=1

n∈N

nÎN

6r = 3k(k +1)r=1

k

∑

6r=3k(k+1)

r=1

k

å

n = k +1

n=k+1

n∈N

nÎN

6r = 6r + 6 × (k +1)r=1

k

∑r=1

k+1

∑ = 3k(k +1)+ 6(k +1) = (k +1)(3k + 6) = 3(k +1)(k + 2)

6r=6r+6´(k+1)

r=1

k

å

r=1

k+1

å

=3k(k+1)+6(k+1)=(k+1)(3k+6)=3(k+1)(k+2)

3k(k +1)⇒ 3(k +1)(k +1+1) = 3(k +1)(k + 2)

3k(k+1)Þ3(k+1)(k+1+1)=3(k+1)(k+2)

n = k +1

n=k+1

6r = 3n(n +1)r=1

n

∑

6r=3n(n+1)

r=1

n

å

n = 1

n=1

1.2

Applying summation formulae

·

use mathematical induction to prove summation formulae

1

.

Use proof by induction to show that, for all

n

³

1

n

³

1

,

n

Î

N

n

Î

N

(a)

r

r

=

1

n

å

=

n

(

n

+

1

)

2

r

r

=

1

n

å

=

n

(

n

+

1

)

2

(b)

6

r

r

=

1

n

å

=

3

n

(

n

+

1

)

6

r

r

=

1

n

å

=

3

n

(

n

+

1

)

(c)

8

r

r

=

1

n

å

=

4

n

(

n

+

1

)

8

r

r

=

1

n

å

=

4

n

(

n

+

1

)

(d)

2

r

-

1

=

n

2

r

=

1

n

å

2

r

-

1

=

n

2

r

=

1

n

å

(e)

(

6

r

-

1

)

r

=

1

n

å

=

n

(

3

n

+

2

)

(

6

r

-

1

)

r

=

1

n

å

=

n

(

3

n

+

2

)

(f)

8

r

-

5

=

n

(

4

n

-

1

)

r

=

1

n

å

8

r

-

5

=

n

(

4

n

-

1

)

r

=

1

n

å

1.5

Applying

algebraic and geometric skills to methods of proof

·

Disproving a con

jecture by providing a counter

-

example

2

.

(a)

For any real number

s

a

,

b

,

c

a

,

b

,

c

and

d

d

, it is conjectured that

a

<

b

a

<

b

and

c

<

d

Þ

ac

<

bd

c

<

d

Þ

ac

<

bd

Find a counter

-

example to disprove this conjecture.

(b)

For any real numbers

a

a

and

b

b

it is conjectured that

a

b

<

1

Þ

a

<

b

a

b

<

1

Þ

a

<

b

Find a counter

-

example to

disprove this conjecture.

(c)

For any real numbers

a

a

and

b

b

it is conjectured that

a

2

>

b

2

Þ

a

>

b

a

2

>

b

2

Þ

a

>

b

Find a counter

-

example to disprove this conjecture.

1.5

Applying

algebraic and

geometric skills to methods of proof

·

Using direct and indirect proof in straightforward examples

3

.

(a)

Prove by contradiction, that if

3

n

3

n

is even then

n

n

is also even, where

n

n


(b)


5

n

5

n

is even then

n

n

is also even, where

n

n


(c)


n

2

n

2

is odd then

n

n

is also odd, where

n

n


(d)


2

n

2

2

n

2

is odd then

n

n

is also odd, where

n

n


(

e

)

Use direct proof to

show that

if 3

is added to the square of

an

odd inte

ger

the answer is divisible by 6

.

(

f)

Use direct proof to

show that

the product

of

an

y

two

odd inte

ger is an even integer.

(g)

Use direct proof to

show that

if

5 is taken away from any odd number the answer is a

multiple of

2

.

1.2 Applying summation formulae

use mathematical induction to prove summation formulae

1. Use proof by induction to show that, for all n1n1, nNnN

(a) r

r1

n

n(n1)

2

r

r1

n

n(n1)

2

(b) 6r

r1

n

3n(n1)6r

r1

n

3n(n1) (c) 8r

r1

n

4n(n1)8r

r1

n

4n(n1)

(d) 2r1n

2

r1

n

2r1n

2

r1

n

(e) (6r1)

r1

n

n(3n2)(6r1)

r1

n

n(3n2) (f) 8r5n(4n1)

r1

n

8r5n(4n1)

r1

n


Disproving a conjecture by providing a counter-example

2. (a) For any real numbers

a,b,ca,b,c

and dd, it is conjectured that

abab and cdacbdcdacbd


(b) For any real numbers

aa

and bb it is conjectured that

a

b

1ab

a

b

1ab


(c) For any real numbers

aa

and bb it is conjectured that

a

2

b

2

aba

2

b

2

ab



Using direct and indirect proof in straightforward examples

3. (a) Prove by contradiction, that if 3n3n is even then

nn

is also even, where

nn


(b) Prove by contradiction, that if 5n5n is even then

nn

is also even, where

nn


(c) Prove by contradiction, that if

n

2

n

2

is odd then

nn

is also odd, where

nn


(d) Prove by contradiction, that if

2n

2

2n

2

is odd then

nn

is also odd, where

nn


(e) Use direct proof to show that if 3 is added to the square of an odd integer

the answer is divisible by 6.

(f) Use direct proof to show that the product of any two odd integer is an even integer.

(g) Use direct proof to show that if 5 is taken away from any odd number the answer is a

multiple of 2.

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