1.1The Principle of Mathematical Induction 1.2Divisibility Chapter Summary Case Study Mathematical...

1.1 The Principle of Mathematical Induction

1.2 Divisibility

Chapter Summary

Case Study

Mathematical Induction1

P. 2

In a supermarket, some cans of coke are being piled up such that the top layer has 1 can, the second layer has 1 + 2 3 cans, the third layer has 1 2 3 6 cans, and so on.

Case StudyCase Study

The manager of the supermarket suggests that there should be a total

of n(n 1)(n 2) cans in n layers.6

1

Do you think that his formula is correct? If yes, how can you prove it?

That’s easy!We can find a formula to calculate it. Let me explain it to you.

How can I find the total number of cans in all layers?

P. 3

Consider the following sums:

1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction

S(1)

S(2) 3

2

32

1

21

1

S(3) 4

3

43

1

32

1

21

1

S(4) 5

4

54

1

43

1

32

1

21

1

2

1

21

1

We can guess that the result of S(n):1)1(

1...

32

1

21

1

n

n

nn

However, to prove that S(n)

is true, we have to check whether the formula is true for every positive integer n, which is impossible to do so!

1)1(

1...

32

1

21

1

n

n

nn

P. 4

Instead, we can use a simple method called mathematical induction to prove that this kind of propositions is true for all positive integers n.


Principle of Mathematical InductionLet P(n) be a proposition that involves a positive integer n. Then P(n) is true for all positive integers n if both of the following conditions are satisfied: (I) P(1) is true. (II) For any positive integer k, if P(k) is assumed to be true,

then P(k 1) is also true.

Now, let us apply Mathematical Induction to prove some propositions as illustrated in Example 1.1T, 1.2T and 1.3T.

In other words, we only need to check two conditions, (I) and (II), in order to show that a proposition P(n) is true for all positive integers n.

P. 5


Example 1.1TProve, by mathematical induction, that

for all positive integers n.

Solution:For n 1,

)2)(1(3

1)1(...433221 nnnnn

L.H.S. 1 2 2 R.H.S. 2)21)(11)(1(3

1

∴ The proposition is true for n 1.

Next, assume the proposition is true for some positive integers k, that is,

)2)(1(3

1)1(433221 kkkkk

When n k 1,L.H.S. 1 2 2 3 3 4 k(k 1) (k 1)(k 2) )2)(1()2)(1(

3

1 kkkkk Use the assumption

P. 6


Example 1.1TProve, by mathematical induction, that


Solution:

)2)(1(3

1)1(...433221 nnnnn

When n k 1,L.H.S. 1 2 2 3 3 4 k(k 1) (k 1)(k 2)

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

1 kkkkk

When n k 1, prove that

L.H.S. .)3)(2)(1(31 kkk

]2)1][(1)1)[(1(3

1 kkk

)3)(2)(1(3

1 kkk

∴ The proposition is true for n k 1.

R.H.S.

33

1

∴ By the principle of mathematical induction, the proposition istrue for all positive integers n.

Take out the common factor )2)(1(31 kk

P. 7


Example 1.2T

Solution:For n 1,

L.H.S. R.H.S.

(a) Prove, by mathematical induction, that

for all positive integers n.46)23)(13(

1

118

1

85

1

52

1

n

n

nn

10

1

52

1

10

1

4)1(6

1



46)23)(13(

1

118

1

85

1

52

1

k

k

kk

When n k 1,

L.H.S. )23)(13(

1

118

1

85

1

52

1

kk

)53)(23(

1

kk

P. 8


Example 1.2T(a) Prove, by mathematical induction, that


Solution:

46)23)(13(

1

118

1

85

1

52

1

n

n

nn

When n k 1,

L.H.S. )23)(13(

1

118

1

85

1

52

1

kk

)53)(23(

1

kk

)53)(23(2

)1)(23(

kk

kk

)53)(23(2

253 2

kk

kk)53)(23(2

2)53(

kk

kk)53)(23(

1

46

kkk

k

4)1(6

1

k

k106

1

k

k

R.H.S.


P. 9


Example 1.2T

1798

11

4)9(6

9

4)20(6

20

2926

1

85

1

52

1

6259

1

85

1

52

1

6259

1

3532

1

3229

1



(b) Hence find the value of .

Solution:

46)23)(13(

1

118

1

85

1

52

1

n

n

nn

6259

1

3532

1

3229

1

n 20 n 9

P. 10

Remarks:Although mathematical induction is a very good method for proving a proposition, it may NOT be the only method in some cases.


For instance, the proposition in Follow-up 1.2

can also be proved by using the following fact:

12

1

12

1

2

1

)12)(12(

1

kkkk

12)12()12(

1...

53

1

31

1

n

n

nn

P. 11


Example 1.3T

Solution:

3

)14)(1()2)(12(654321

nnnnn

For n 1,L.H.S. 1 2 2 R.H.S. 2)14)(11)(1(

3

1



When n k 1,L.H.S. 1 2 3 4 5 6 (2k 1)(2k) (2k 1)(2k 2)

3

)14)(1()2)(12(654321

kkkkk

)1)(12(23

)14)(1( kkkkk



P. 12


Example 1.3T

Solution:

3

)14)(1()2)(12(654321

nnnnn

When n k 1,L.H.S. 1 2 3 4 5 6 (2k 1)(2k) (2k 1)(2k 2)

)1)(12(23

)14)(1( kkkkk

3

)34)(2)(1( kkk3

)6114)(1( 2 kkk

)]12)(2)(3()14([3

)1(

kkk

k

3

]1)1(4][1)1)[(1( kkk R.H.S.




P. 13


Example 1.3T

Solution:

3

)14)(1()2)(12(654321

nnnnn

600 166

)50)(49(33

)195)(50(49

)98642(3)9897654321(

98100684624 10098866442

98)397(6)35(4)33(2)31(

n 49 n 49


for all positive integers n.(b) Hence evaluate 2 4 4 6 6 8 ... 98 100. [Hint: You may

use the fact 2 4 6 ... 2n n(n 1) without proof.]

P. 14

Apart from proving propositions that involve summation,mathematical induction can also be used in many other situations.

1.2 Divisibility1.2 Divisibility

For example, 28 is divisible by 4 as 28 4 7.

Suppose p and q are two integers, where q 0. If p qm, where m is an integer, then p is divisible by q.

P. 15


Example 1.4TProve, by mathematical induction, that 11n – 4n is divisible by 7 for all positive integers n.

Solution:For n 1,

11n – 4n 11 – 4 7, which is divisible by 7.


Next, assume for some positive integers k, 11k – 4k is divisible by 7, that is, 11k – 4k 7m, where m is a positive integer.When n k 1,

)4(4)11(11411 1 1 kkkk

)4(777 km )4(4)4(1177 kkm

)4(4)47(11 kkm )411(7 km

∴ 11k + 1 – 4k + 1 is divisible by 7. ∴ The proposition is true for n k 1.

P. 16


Example 1.5TProve, by mathematical induction, that 3 52n 1 23n 1 is divisible by 17 for all positive integers n.

Solution:For n 1,

3 52n 1 23n 1 3 53 24

391, which is divisible by 17. ∴ The proposition is true for n 1.

Next, assume for some positive integers k, 3 52k 1 23k 1 is divisible by 17, that is, 3 52k 1 23k 1 17m, where m is a positive integer.When n k 1,

1 )1 (31 )1 (2 253 kk

)225(17 1 3 km)2(8)2(25)17(25 1 31 3 kkm

8225)217( 1 31 3 kkm

31 321 2 22553 kk

4 33 2 253 kk

∴ 3 52(k 1) 1 23(k 1) 1 is divisible by 17. ∴ The proposition is true for n k 1.

P. 17


Example 1.6TProve, by mathematical induction, that 6n 1 – 5(n 1) – 1 is divisible by 25 for all positive integers n.

Solution:For n 1,

6n 1 – 5(n 1) – 1 62 – 5(2) – 1 25, which is divisible by 25. ∴ The proposition is true for n 1.

Next, assume for some positive integers k, 6k 1 – 5(k 1) – 1 is divisible by 25, that is, 6k 1 – 5(k 1) – 1 25m, where m is a positive integer.When n k 1,

1]1)1[(56 1 )1 ( kk

)16(25 km6)1(56)1(30)25(6 kkm6)1(56]1)1(525[ kkm

15)1(566 1 kk ∴ 6(k 1) 1 – 5[(k 1) 1] – 1is divisible by 25.

∴ The proposition is true forn k 1.

P. 18

1.1 Principle of Mathematical Induction

Chapter Chapter SummarySummary

A proposition P(n) is true for all positive integers n if both of the following conditions are satisfied: (I) P(1) is true. (II) For any positive integer k, if P(k) is assumed to be true, then

P(k 1) is also true.

P. 19

1.2 Divisibility

Chapter Chapter SummarySummary

Suppose p and q are two integers. p is divisible by q (q 0) if p qm, where m is an integer.

1.1The Principle of Mathematical Induction 1.2Divisibility Chapter Summary Case Study Mathematical...

Documents

Transcript of 1.1The Principle of Mathematical Induction 1.2Divisibility Chapter Summary Case Study Mathematical...