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Mathematics Extension 1 Syllabus: 7.4 Mathematical Induction. Applications Mathematical Induction A compilation of material from Stuart Palmer and Pauline Hunt. ‘Practising Induction Prerequisities’ © Sue Thomson, Ian Forster

Mathematical Induction Booklet

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Page 1: Mathematical Induction Booklet

Mathematics Extension 1 Syllabus: 7.4 Mathematical Induction. Applications

Mathematical Induction

A compilation of material from Stuart Palmer and Pauline Hunt. ‘Practising Induction Prerequisities’ © Sue Thomson, Ian Forster

Page 2: Mathematical Induction Booklet

16

Practising Indudion Prerequisites

• Show Ihese expres_ ions are true.

(i) 2K 3 = 2(K I)

(ii) 3K + ~ = 3(K I) + I

(iii) 5K 7 = 5(K + I) 2

(iv) K - I = (K I) - 2

(v) 5K + ~ = 5(K 11) - 1

.. Complele lhese expressions.

(i) K+3 = (K+I) _ __ _

(ii) 2K 5 = 2(K + I) ___ _

(vii) 2 X 2K = 2-­

(viii) 6K X 6 = 6--

(iii) 7K + 11 = 7(K + 1) ___ _

(iv) K - 3 = (K + I) ___ _

(v) 2K - 1 = 2(K + 1) _ __ _

(vi) 5K - 1 = 5(K + I) _ __ _

(ix) 2K + 2K = _ _ X 2K

- 2 -

(x) 3K + 3A- 3K =

= 3

• Show these expres ions are true by faclorising the left hand side (LHS).

1 Hinl. remove a fraclion and a binomial faclor.)

I I (i) "jK(K + I) + (K -t I)(K + 2) = /K i 1)(4K 6

(ii) iK(K - I) + K(K + I) = ~K(3K + I)

(iii) ~(K + I) + (K I)(K + 2) = 1K + 1)(7K + 8)

.. Provc the. e idenlilie. truc.

(i) 2K(K - 2) + 4(K + I) - 6 = 2(K I)[(K I) - 2J

(ii) jK(K + I)(K + 2) + (K + I)(K + 2) = !(K + I)(K + 2)(K + 3)

(iii) ~K(3K I) KI3(K + I) - 11

K K

(iv) (K + I) + (K + IH(K + I) I]

! K(9K + 5)

(K I

(K + I)

K(K+I) (K'+lr (K+I)I(K+I)+I]

(v) 2(2K I)" [2(K + I) - 1J[2 K + I) + IJ ~ 212(K + I) I1

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Page 3: Mathematical Induction Booklet

Mathematical Induction (Extension 1)

Warm-up activity:

Calculation 1 1

1 2!

2 1 1

1 2 2 3+

! !

3 1 1 1

1 2 2 3 3 4+ +

! ! !

4 1 1 1 1

1 2 2 3 3 4 4 5+ + +

! ! ! !

5 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6+ + + +

! ! ! ! !

6 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7+ + + + +

! ! ! ! ! !

7 1 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7 7 8+ + + + + +

! ! ! ! ! ! !

8 1 1 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9+ + + + + + +

! ! ! ! ! ! ! !

etc etc n?

In this topic, you will learn a method of proof, called mathematical induction, with which you can prove that the sum of an infinite pattern is equal to a simple expression.

Page 4: Mathematical Induction Booklet

Mathematical Induction Syllabus: 7.4 Mathematical Induction. Applications Mathematical Induction is a method of proof that can be used to prove that a statement IS true for ALL positive integers n. There are 3 main types of induction questions:

1. Proving a summation statement: Prove that 1+ 3+5+ ...+ (2n−1) = n2 , for all positive integers n.

2. Proving a divisibility statement:

Prove that 5n −1 is divisible by 4, for all positive integers n.

3. Proving an inequality statement: Prove that 2n > n2 , for n > 4 .

How to do a proof by induction Step 1: Show that the statement is true for the smallest value of n (usually n =1)

1st Prove that the first budgie is blue. Substitute n = 1

Step 2: Assume that the statement is true for n = k

kth Assume it is blue. Replace n with k.

Step 3: Prove that the statement is true for n = k +1

Check all budgies tell the truth.

Step 4: (The conclusion) If the result is true for n = k, then it is also true for n = k +1. Since it is true for n = 1, it must also be true for n = 2, n = 3 and so on for all positive integers n. or Hence the statement is true for all integers n ≥1 , by induction.

We have assumed the kth budgie is blue, then proved that the (k + 1)th budgie (i.e. the next one in line) is also blue. Since the first one is blue, so is the second and therefore so is the third and the fourth so on.

Page 5: Mathematical Induction Booklet

WORKSHEETS ON MATHEMATICAL INDUCTION

Example 1: TYPE 1 SUMMATION Your question

Show by induction that :

for n a positive integer

Show by induction that :

for n a positive integer

step 1: Test n = 1

step 2: Assume true for n = k

step 3: Prove result is true for n = k + 1

step 4: True for all positive integers n≥ 1 by induction

12 + 22 + 32 + ....+ n2 = 16n(n +1)(2n +1)

1+ 2 + 3+ ....+ n = n2(n +1)

Page 6: Mathematical Induction Booklet

Example 2: TYPE 1 SUMMATION Your question

Show by induction that :

for n a positive integer

Show by induction that :

for n a positive integer

step 1: Test n = 1 LHS: Tn = T1 = 1 .2.3 =

RHS:

= 6 = LHS

True for n = 1

step 2: Assume true for n = k

step 3: Assume true for n = k + 1 ie Sk+1 =

=

LHS:

becomes

(This is just Sk + Tk+1) Factorise (k+1) (k+2) (k+3) from both terms

=

since then take out factor of

=

If the result is true for n = k then it is true for n = k+1

step 4: True for all positive integers n≥ 1 by induction

1.2.3+ 2.3.4 + ....+ n(n +1)(n + 2) = n4(n +1)(n + 2)(n + 3) 1.2 + 2.3+ 3.4....+ n(n +1) = n

3(n +1)(n + 2)

n(n +1)(n + 2)

Sn =n4(n +1)(n + 2)(n + 3)

S1 =14(1+1)(1+ 2)(1+ 3)

1.2.3+ 2.3.4 + ....+ k(k +1)(k + 2) = k4(k +1)(k + 2)(k + 3)

1.2.3+ 2.3.4 + ....+ k(k +1)(k + 2)+ (k +1)(k + 2)(k + 3)

= (k +1)4

(k +1+1)(k +1+ 2)(k +1+ 3)

= 14(k +1)(k + 2)(k + 3)(k + 4)

1.2.3+ 2.3.4 + ....+ k(k +1)(k + 2)+ (k +1)(k + 2)(k + 3)

k4(k +1)(k + 2)(k + 3)+ (k +1)(k + 2)(k + 3)

(k +1)(k + 2)(k + 3) k4+1

⎡⎣⎢

⎤⎦⎥

k4+1

⎡⎣⎢

⎤⎦⎥= k4+ 44

⎡⎣⎢

⎤⎦⎥

14

14(k +1)(k + 2)(k + 3) k + 4[ ]

Page 7: Mathematical Induction Booklet

Example 3: TYPE 1 SUMMATION Your question

Show by induction that :

for n a positive integer

Show by induction that :

for n a positive integer

step 1: Test n = 1

LHS: Tn = ; T1 =

=

RHS:

= = LHS

True for n = 1

step 2: Assume true for n = k

step 3: Assume true for n = k + 1

LHS =

=

=

=

=

= RHS If the result is true for n = k then it is true for n = k+1

step 4: True for all positive integers n≥ 1 by induction

21.4

+ 24.7

+ ......+ 1(3n − 2)(3n +1)

=1− n3n +1

21.3

+ 23.5

+ ......+ 2(2n −1)(2n +1)

=1− 12n +1

1(3n − 2)(3n +1)

1(3− 2)(3+1)14

Sn =1−n

3n +1

S1 =1−13+1

14

21.4

+ 24.7

+ ......+ 1(3k − 2)(3k +1)

=1− k3k +1

21.4

+ 24.7

+ ......+ 1(3k − 2)(3k +1)

+ 1(3 k +1( ) − 2)(3 k +1( ) +1)

=1− k +13 k +1( ) +1

21.4

+ 24.7

+ ......+ 1(3k − 2)(3k +1)

⎣⎢

⎦⎥+

1(3k +1)(3k + 4)

=1− k +13k + 4

1− k3k +1

⎡⎣⎢

⎤⎦⎥+ 1(3k +1)(3k + 4)

1− k(3k + 4)(3k +1)(3k + 4)

+ 1(3k +1)(3k + 4)

1− 3k2 + 4k +1(3k +1)(3k + 4)

1− (3k +1)(k +1)(3k +1)(3k + 4)

1− k +13k + 4

Page 8: Mathematical Induction Booklet

Type 1: Summation Additional Questions: example 1: Prove by induction for n a positive integer:

a)

b)

c)

d) 1+ 2+ 22 + ...+ 2n−1 = 2n −1 example 2: Prove by induction for n a positive integer:

a)

b) 1.2+ 2.4+ 3.5+ ...+ n(n+ 2) = n

6(n+1)(2n+ 7)

example 3: Prove by induction for n a positive integer:

a)

b)

c) x≠ 0 or 1

13 + 23 + 33 + ....+ n3 = 14n2 (n +1)2

1+ 3+ 5 + ....+ (2n −1) = n2

12 + 32 + 52 + ....+ (2n −1)2 = 13n(2n −1)(2n +1)

13 + 23 + 33 + ....+ n3 = 14n2 (n +1)2

11.3

+ 13.5

+ ......+ 1(2n −1)(2n +1)

= n2n +1( )

11.2

+ 12.3

+ ......+ 1n(n +1)

=1− 1n +1

1x −1

− 1x− 1x2

− 1x3......− 1

xn= 1xn x −1( )

Page 9: Mathematical Induction Booklet

Example 4: TYPE 2 DIVISIBILITY Your question

Show by induction that : is divisible by 4 for n a positive integer

Show by induction that : is divisible by 8 for n a positive integer

step 1: Test n = 1 when n = 1 = = 4 which is divisible by 4 True for n = 1

step 2: Assume true for n = k = 4 x P where P is any integer or = 4P + 1

step 3: Prove result is true for n = k + 1 is divisible by 4 = 4 x Q where Q is an integer LHS: = = ( 4P + 1)5 -1 = 20P + 5 - 1 = 20P - 4 = 4 ( 5P - 1) = 4 x Q which is divisible by 4 If the result is true for n = k then it is true for n = k+1

step 4: True for all positive integers n≥ 1 by induction

Type 2: Divisibility Additional Questions: example 4: Prove by induction a) if n is even then n2 + 2n is divisible by 8 b) if n is even then 9n - 3 is a multiple of 6

5n −1 32n −1

5n −1 51 −1

5k −15k

5k+1 −15k+1 −1

5k+1 −1 5k.51 −1

Page 10: Mathematical Induction Booklet

Example 5: TYPE 2 DIVISIBILITY Your question

Show by induction that : is divisible by (x-1) for n a positive integer

Show by induction that : is divisible by (a-1) for n a positive integer

step 1: Test n = 1

when n = 1: x - 1 is divisible by (x-1) (itself)

True for n = 1

step 2: Assume true for n = k = where P is any polynomial

step 3: Prove result is true for n = k + 1 is divisible by (x - 1) ie: = (x-1).Q where Q is any polynomial LHS: =

=

=

=

=

=

=

which is divisible by (x-1)

If the result is true for n = k then it is true for n = k+1

step 4: True for all positive integers n≥ 1 by induction

xn −1 an −1

xk −1 x −1( )P

xk+1 −1xk+1 −1

xk+1 −1 xk .x1 −1

xk+1 − x + x −1

xk .x1 − x + x −1

x xk −1( ) +1 x −1( )x x −1( )P +1 x −1( )x −1( ) Px +1( )x −1( )Q

Page 11: Mathematical Induction Booklet

Example 6: TYPE 3 INEQUALITIES Your question

Show by induction that for n a positive integer

Show by induction that : for n a positive integer

step 1: Test n = 1

when n = 1:

True for n = 1

step 2: Assume true for n = k

step 3: Prove result is true for n = k + 1 show ie: *** LHS: from step 2:

*** break 6k up into 2k +4k

since k is a positive integer

If the result is true for n = k then it is true for n = k+1

step 4: True for all positive integers n≥ 1 by induction

Type 3: Inequality Additional Questions: example 6: Prove by induction where p > -1

3n ≥1+ 2n 5n ≥1+ 4n

31 ≥1+ 2

31 ≥ 3

3k ≥1+ 2k

3k+1 ≥1+ 2(k +1)

3k+1 ≥ 2k + 3

3k+1 = 31.3k

3k ≥1+ 2k31.3k ≥ 3(1+ 2k)

31.3k ≥ 3+ 6k

3k+1 ≥ 2k + 3+ 4k

3k+1 ≥ 2k + 3+ 4k 2k + 3

(1+ p)n ≥1+ np

Page 12: Mathematical Induction Booklet

Additional Exercises: Prove the following, using mathematical induction, for all positive integers n (except when asked to do otherwise):

S1: 1

( 1)2

n

r

nr n

=

= +!

S2: 1

(3 1) (3 1)2

n

r

nr n

=

! = +!

S3: 1

1

2 2 1n

r n

r

!

=

= !!

S4: 1

1

3(4 ) 4 1n

r n

r

!

=

= !!

S5: 2

1

( 1)(2 1)6

n

r

nr n n

=

= + +!

S6: 2

3 2

1

( 1)4

n

r

nr n

=

= +!

S7: 1 1 1

..........1 2 2 3 ( 1) 1

n

n n n+ + + =

" " + +

S8: 1 1 1 1

..........1 4 4 7 7 10 (3 2)(3 1) 3 1

n

n n n+ + + + =

" " " ! + +

S9: 2

1

(2 1)n

r

r n=

! =!

S10: 1 2 3 2 2 1.....n n

n n n n n x cx x c x c xc c

x c

! ! ! ! ! !+ + + + + =

!

S11: [ ]1

( 1) 2 ( 1)2

n

r

na r d a n d

=

+ ! = + !!

S12: 1

1

(1 )

(1 )

nni

i

a rar

r

!

=

!=

!!

S13: 2 2 2 21 .2 2 .3 3 .4 ......... .( 1) ( 1)( 2)(3 1)12

nn n n n n+ + + + + = + + +

Page 13: Mathematical Induction Booklet

S14: 2 2

1

(2 1) (4 1)3

n

r

nr n

=

! = !!

S15: 1

( 1)( 2)( 3)( 1)( 2)

4

n

r

n n n nr r r

=

+ + ++ + =!

S16: 3 3 3 3 2 22 4 6 ........ (2 ) 2 ( 1)n n n+ + + + = +

S17: 1

( 1)( 2)(1 )

3

n

r

n n nr r

=

+ ++ =!

S18: 1

( 1)(2 7)( 2)

6

n

r

n n nr r

=

+ ++ =!

S19: 1

(3 1)(3 2)

2

n

r

n nr

=

!! =!

S20: 1

1

(4 3)(4 1) 4 1

n

r

n

r r n=

=! + +

!

S21: 1

1log log( 1)

n

r

rn

r=

+= +!

S22: 1

( !) ( 1)! 1n

r

r r n=

= + !!

S23: 1

( 1) 1!

( 1)! ( 1)!

n

r

r n

r n=

+ !=

+ +!

S24: 5 6 7 4 (3 7)

..........1 2 3 2 3 4 3 4 5 ( 1)( 2) 2( 1)( 2)

n n n

n n n n n

+ ++ + + + =

" " " " " " + + + +

S25: 0 1 2 3 12.2 3.2 4.2 5.2 ..... ( 1)2 .2n nn n!+ + + + + + =

S26: 1

1

.2 ( 1)2 2n

r n

r

r n +

=

= ! +!

S27: 1

( 1) ( 1)( 2)

2 6

n

i

i i n n n

=

+ + +=!

S28: ( )1

1

2 1 (2 1) 2 1

n

j

n

j j n=

=! + +

!

Page 14: Mathematical Induction Booklet

D1: 9 7n n! is even D2: ( 1)( 2)n n n+ + is divisible by 3

D3: ( 1)n n + is even

D4: 7 3n n! is divisible by 4

D5: 29 4n n+ ! is a multiple of 5

D6: 3 2n n+ is divisible by 3

D7: 8 5n n! is divisible by 3 D8: ( 2)n n + is divisible by 8 if n is even

D9: 3 2n n+ is a multiple of 5 when n is odd

D10: 5 1n ! is divisible by 4 D11: ( 1)(2 1)n n n+ + is divisible by 6

D12: 7 2n + is divisible by 3

D13: 7 3 .7 1n nn+ ! is divisible by 9

D14. 2 2n nx y! is divisible by x y+

D15: 19 8 9n n+ ! ! is divisible by 64

D16: 23 1n ! is divisible by 8

D17: 43 1n ! is divisible by 80

D18: 2 2n n+ is a multiple of 8 if n is even

D19: 27 48 1n n! ! is divisible by 2304

D20: 7 19n n+ is a multiple of 13 when n is odd For Extension 2 thrill-seekers:

D21: 5n n! is divisible by 30

Page 15: Mathematical Induction Booklet

I1: 3n n>

I2: 3 2n n>

I3: 22n n> , for 4n >

I4: 23n n>

I5: 3 1 2n n! +

I6: (1 ) 1np np+ ! +

I7: 4 1 3n n! + , for 1n >

I8: 2 22 1n n n! + + , for 1n >

I9: 2!n n> , for 4n !

I10: ! 2nn > , for 4n >

I11: 2

4 (2 )!

1 ( !)

n n

n n"

+, for 1n >

I12: 55n n! , for 5n !

I13: ! 3nn > , for 7n !

Induction HSC Questions (Extension 1) Year Quest Marks Content

2008 3b 3 Summation 2007 4b 3 Divisibility 2006 5dii 3 Summation 2005 4d 3 Inequality 2004 4a 3 2003 3d 3 Summation 2002 5a 3 Summation 2001 6a 3 Divisibility 2000 4a 3 Summation 1999 5a 3 1998 3a 3 Divisibility 1997 5b 5 Summation 1994 3c Divisibility 1993 5a Summation 1992 4b Summation

Page 16: Mathematical Induction Booklet

Mathematical Induction (Extension 1)

Warm-up activity:

Calculation 1 1

1 2!

2 1 1

1 2 2 3+

! !

3 1 1 1

1 2 2 3 3 4+ +

! ! !

4 1 1 1 1

1 2 2 3 3 4 4 5+ + +

! ! ! !

5 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6+ + + +

! ! ! ! !

6 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7+ + + + +

! ! ! ! ! !

7 1 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7 7 8+ + + + + +

! ! ! ! ! ! !

8 1 1 1 1 1 1 1 1

1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9+ + + + + + +

! ! ! ! ! ! ! !

etc etc n?

In this topic, you will learn a method of proof, called mathematical induction, with which you can prove that the sum of an infinite pattern is equal to a simple expression.