Revision Proof by Induction

7/28/2019 Revision Proof by Induction

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Further Pure 1

Revision Topic 3: Proof by induction

The OCR syllabus says that candidates should be able to:(a) use the method of mathematical induction to establish a given result (not restricted to

summation of series);

(b) recognise situations where conjecture based on a limited trial followed by inductive proof is a

useful strategy, and carry this out in simple cases, e.g. to find the nth power of the matrix1 1

0 1

.

Section 1: General principals

There are two steps involved in proving a result by induction:

Step 1: Prove true when n = 1.

Step 2: (The inductive step). Assume the result is true forn= kand then prove true forn = k+ 1.

1.1 Summing series

Example: Prove by induction that2

61

( 1)(2 1)n

n

r

r n n=

= + + for all positive integer values ofn.

Solution: We wish to show that2 2 2 2

61 2 3 ... ( 1)(2 1)nn n n+ + + + = + + (*)

Step 1: This is to prove the result true when n = 1:

Left hand side of equation (*) = 12 = 1.

Right hand side of equation (*) =16

(1 1)(2 1)= + + = 1.

So equation (*) is true when n = 1.

Step 2: We assume the result is true when n = k, i.e. we assume that

2 2 2 2

61 2 3 ... ( 1)(2 1)kk k k+ + + + = + +

We want to prove the result is true when n = k + 1, i.e. we wish to show that

2 2 2 2 2 1

6

1 2 3 ... ( 1) ( 1 1)(2( 1) 1)kk k k k ++ + + + + + = + + + +

i.e.2 2 2 2 2 1

61 2 3 ... ( 1) ( 2)(2 3)kk k k k ++ + + + + + = + + .

But2 2 2 2 2 2

61 2 3 ... ( 1) ( 1)(2 1) ( 1)kk k k k k + + + + + + = + + + + (using our assumption).

So, [ ]( 1)2 2 2 2 2

61 2 3 ... ( 1) (2 1) 6( 1)

kk k k k k

++ + + + + + = + + +

Therefore,( 1) ( 1)2 2 2 2 2 2 2

6 61 2 3 ... ( 1) 2 6 6 (2 7 6)

k kk k k k k k k

+ + + + + + + + = + + + = + +

Factorising we get:2 2 2 2 2 1

6

1 2 3 ... ( 1) ( 2)(2 3)kk k k k ++ + + + + + = + + as required.

Therefore the result is true forn = k+ 1.


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So, by induction, the result is true for all integers n 1.

Examination style question

Prove by induction that 21

( 1)n

n

r

r n=

= + for all positive integer values ofn.

Worked examination question (AQA January 2006)

a) Prove by induction that

2 12 (3 2) (4 2 ) ... ( 1)2 2n nn n+ + + + + = (*)

for all integers n 1.

b) Show that

21 1

1

( 1)2 2 (2 1)n

r n n

r n

r n +

= ++ =

Solution:

Step 1: This is to prove the result true when n = 1:


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Left hand side of equation (*) = 1 1(1 1)2 2+ =

Right hand side of equation (*) = 1 21 = 2.

So equation (*) is true when n = 1.

Step 2: We assume the result is true when n = k, i.e. we assume that

2 12 (3 2) (4 2 ) ... ( 1)2 2k kk k+ + + + + = .

We want to prove the result true when n = k+ 1, i.e. we want to show that

2 1 1 12 (3 2) (4 2 ) ... ( 1 1)2 ( 1)2k kk k+ ++ + + + + + = +

i.e. that 2 1 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 ( 1)2k k kk k k ++ + + + + + + = + .

But, 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( 2)2k k k k k k k k + + + + + + + = + + (using the assumption)

So 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( ( 2))k k kk k k k + + + + + + + = + +

i.e. 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 (2 2)k k kk k k+ + + + + + + = +

i.e. 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 2( 1)k k kk k k+ + + + + + + = +

So we have 2 1 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( 1)k k kk k k ++ + + + + + + = + as required.

Therefore the equation is true when n = k+ 1.

So the equation is true for all integer values n 1.

b)2 2

1 1 1

1 1 1

( 1)2 ( 1)2 ( 1)2n n n

r r r

r n r r

r r r

= + = =

+ = + +

Using the result from part (a), we know that2

1 2

1

( 1)2 2 2n

r n

r

r n

=+ =

and1

1

( 1)2 2n

r n

r

r n

=+ = .

Therefore,2

1 2

1

( 1)2 2 2 2 2 2 2 2n

r n n n n n

r n

r n n n n

= ++ = =

Factorising gives:

21 1

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

nr n n n n

r nr n n

+

= + + = = as required.

Examination question (OCR January 2005)

Prove by induction that13

1 4 2 5 3 6 ... ( 3) ( 1)( 5)n n n n n + + + + + = + +

for all integers n 1.


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Examination question (Edexcel)

Prove, by induction, that

2 112

1

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

r

r r n n n n=

= + +


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1.2 Sequences

Example: Prove that if 1 3 4 for n nu u n+ = + , and 1 2u = , then 14 3 2nnu

= .

Solution:

Step 1: We prove the formula true when n = 1: 1 11 4 3 2 4 2 2u= = = (which is true).

Step 2: Assume the formula is true when n = k, i.e. that 14 3 2kku

= .

We need to prove the result true when n = k+ 1, i.e. that 1 1

1 4 3 2 4 3 2

k k

ku

+

+ = = .

But,1

1 3 4 3 4 3 2 4k

k ku u

+ = + = + (using the assumption).

So, 11 3 4 3 6 4 4 3 2k k

ku

+ = + = as required.

Therefore the formula is true when n = k+ 1.

Therefore the result is true for all integers n 1.

Examination question (AQA June 2005)

The sequence 1 2 3, , ,...u u u is defined by

11 1 2

0, ( ).n nu u u n+= = +

Prove by induction that, for all n 1,

( )1

12

2.n

nu n

= +


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Worked examination question (Edexcel 2005)

(a) Express

3

106

+

+

x

xin the formp +

3+x

q, wherep and q are integers to be found.

(1)

The sequence of real numbers u1, u2, u3, ... is such that u1 = 5.2 and un + 1 =3

106

+

+

n

n

u

u

.

(b) Prove by induction that un > 5, forn+.(4)

Solution:

a) Note that p +3+x

q=

( 3)

3

p x q

x

+ +

+

.

Writing3

106

+

+

x

x=

( 3)

3

p x q

x

+ +

+

, we must have 6x + 10 =p(x + 3) + q.

So 6 =p (comparing coefficients ofx)

and 10 = 3p + q i.e. 10 = 18 + q i.e. q = -8.

Therefore3

106

+

+

x

x= 6 -

8

3x +.

b) un + 1 =3

106

+

+

n

n

u

u

= 6 -8

3nu +.

Step 1: Prove the result true when n = 1, i.e. that u1 > 5. This is trivially true as u1 = 5.2.

Step 2: Assume true when n = k, i.e. that uk> 5.

We then need to prove the result true when n = k+ 1, i.e. that uk+1 > 5.

As uk> 5, then uk+ 3 > 8 and so8

3nu +< 1.

Therefore

un + 1 =3

106

+

+

n

n

u

u

= 6 -8

3nu +> 6 1 = 5 (as required)

So the result is true when n = k+ 1.

Therefore the result is true forn+.

Examination question (NICCEA)

Consider the sequence defined by the relationship 1 5 2n nu u+ = + whose first term is 1 1u = .

(i) Show that the first four terms are 1, 7, 37, 187,


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(ii) Use the method of induction to prove that11

23(5 ) 1nnu

= .

1.3 Divisibility

Some questions give you a formula that defines a sequence (in the form un =f(n) ) and then ask

you to prove that all terms of the sequence are divisible by a particular number.

These questions are usually tackled by simplifyingf(k+ 1) f(k) ORf(k+ 1) f(k).

Adapted past examination question (adapted MEI):

Letf(n) = 4 12 3n+ + . By consideringf(n + 1) f(n), or otherwise, prove that 4 12 3n+ + is a

multiple of 5 for any positive integern.

Solution: Letf(n) = 4 12 3n+ + .

Step 1: Provef(1) is a multiple of 5. Butf(1) = 52 3+ = 35 (which is a multiple of 5).

Step 2: We assumef(k) is a multiple of 5. We want to prove thatf(k+ 1) is a multiple of 5.

First we try to simplify:

f(n + 1) -f(n) = 4( 1) 12 3n+ + + - ( 4 12 3n+ + ) = 4 5 4 12 2n n+ + = 4 1 4 4 12 (2 1) 15 2n n+ + =


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Therefore:

f(k+ 1) f(k) = 4 115 2 k+

So, f(k+ 1) = 4 115 2 ( )k f k+ + .

So,f(k+ 1) is a multiple of 5.

Therefore by induction,f(n) must be a multiple of 5 for all positive integers n.

Worked Examination Question: Edexcel 2002

Forn + prove that

23n + 2 + 5n + 1 is divisible by 3,

Solution:

Let f(n) = 23n + 2 + 5n + 1

If we here calculate, f(n + 1) +f(n) = 23(n+ 1) + 2 + 5n+ 1 + 1 + (23n + 2 + 5n + 1)

= 23n + 5 + 5n + 2 + 23n + 2 + 5n + 1

= 23n + 5 + 23n + 2 + 5n + 2 + 5n + 1

= 23n+2(23 + 1) + 5n+1(5 + 1)

= 9 23n+2 + 6 5n+1

Now we are ready to prove the result.

Step 1: First prove true when n = 1:

23 + 2 + 51 + 1 = 32 + 25 = 57 (which is divisible by 3).

Step 2: Assume that the result is true when n = k, i.e. thatf(k) is divisible by 3.

We showed above that f(k+ 1) +f(k) = 9 23k+2 + 6 5k+1

i.e. that f(k+ 1) = 9 23k+2

+ 6 5k+1

f(k)

So f(k+ 1) must be divisible by 3 (as required).

Therefore the result is true forn +

Past examination question (AQA January 2004)

The functionfis given by 3 3 3( ) ( 1) ( 2)f n n n n= + + + + .a) Simplify, as far as possible,f(n + 1) f(n).

Divisible by3 Divisible by3 Divisible by3 (byassumption)

Multiple of 5 Assumed to be a

multiple of 5


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b) Prove by induction that the sum of the cubes of three consecutive positive integers is divisible

by 9.


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Past examination question (Edexcel 2003)

f(n) = (2n + 1)7n 1.

Prove by induction that, for all positive integers n,f(n) is divisible by 4.


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Past examination question (Edexcel)

f(n) = 4 424 2 3n n + , where n is a non-negative integer.

a) Write down f(n + 1) f(n).

b) Prove by induction that f(n) is divisible by 5.


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1.4 Matrix results

Example:

a) Find the matrices

2 31 1 1 1

and0 1 0 1

.

b) Predict the value of1 1

0 1

n

and prove your result true by induction.

Solution

a)2

1 1 1 1 1 1 1 2

0 1 0 1 0 1 0 1

= = 3

1 1 1 2 1 1 1 3

0 1 0 1 0 1 0 1 = =

.

b) It seems sensible to predict that1 1 1

0 1 0 1

nn =

.

We know the result is true when n = 1, 2 and 3.

Suppose now that it is true when n = k, i.e. that

1 1 1

0 1 0 1

kk

= .

We need to prove the result true when n = k+ 1, i.e. that1

1 1 1 1

0 1 0 1

kk

+ + =

.

But:1

1 1 1 1 1 1 1 1 1

0 1 0 1 0 1 0 1 0 1

k kk

+ = =

(by assumption)

Therefore:1

1 1 1 1

0 1 0 1

kk

+ + =

(as required).

So the result is true forn = k+ 1.Therefore our prediction is true for all positive integer values ofn.

Past examination question (MEI adapted)

You are given the matrix1 4

1 3

=

A .

(i) Calculate 2 3andA A .

(ii) Prove by induction that1 2 4

1 2n n n

n n

= + A when n is a positive integer.


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Past examination question (Edexcel 2002)

Prove thatn

49

12=

+

139

31

nn

nn

when n is a positive integer.


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Past examination question

Let1 0

1 2A

= .

Use induction to prove that, for all positive integers n,

21 0

1 2 2

nnA =

.


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1.5 Other examination questions

Past examination question (OCR 2004)

(i) Show that 1 1 ( ) ( )k k k k k x y x x y y x y+ + = + .

(ii) Using this result, prove by induction that (x y) is a factor of n nx y for all integers n 1.

Past examination question (OCR)

Prove by mathematical induction that, for all positive integers n,

1 1 1 1 1... 1

2 4 8 2 2n n

+ + + + =

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