mathematical induction the principle of mathematical induction application in the series application...

Mathematical Induction

The Principle of Mathematical Induction

Application in the Series

Application in divisibility

The Principle of Mathematical Induction

數學歸納法﹝ Mathematical Induction﹞是用來証明某些與

自然數 n 有關的數學命題的一種方法。它的步驟是：

1. 驗証 n = 1 時命題成立﹝這叫歸納的基礎﹞；

2. 假設 n = k 時命題成立﹝這叫歸納假設﹞，

在這假設下証明 n= k+1 時命題成立。

        根據 1、 2 可以斷定命題對一切自然數都 成立。

Application in Series

2n1n2531

Firstly, we need to know the names in the series clearly.

The first term

The n-th term

1n2531 How many terms in the above series?

Can you deduce the (n+1)-th term?

1n2

12n2

11n2

Answer is:

Example 21nn1n3n1037241

Let the proposition is S(n),can you write down S(k)? 21kk1k3k1037241

Also, what is S(k+1)?

k+1

1 2

k

WHO AM I？

211k1k

11k31k1k3k

1037241

In steps of calculation, the meaning of principle of mathematical induction is follows:

nSnnnExample

2

121:

2

1kkk21

Steps means: 2

1kkk21

2

11k1k1kk21

Explanation nSnn

nExample 2

121:

Firstly, prove S(1) is true:

.)1(

12

111..

1..

,1

trueisS

SHR

SHL

nWhen

Assume S(k) is true,

2

1kkk21

Use S(k) is true,prove S(k+1) is also true.

Following steps is the hardest part in the mathematical induction.

2

11k1k1kk21

SHR

kk

kk

kkk

kkk

..2

1112

212

121

12

1

Because S(k) is true

121.. kkSHL

.1 trueiskS By the principle of mathematical induction, S(n) is true for all positive integers n.

Example

3n2n1nn4

12n1nn543432321

Let the proposition is S(n).When n = 1,

.1

6

31211114

1...

6

21111...

trueisS

SHR

SHL

Assume S(k) is true, i.e.,

3k2k1kk4

12k1kk543432321

When n = k+1,

.1

...4

31211114

43214

3214321

3213214

1

32121543432321

trueiskS

SHR

kkkk

kkkk

kkkkkkk

kkkkkkk

kkkkkk

By the principle of mathematical induction, S(n) is true for all positive integers n.

Application in divisibility

The definition of divisibility

0bIf a and b be two integers with

Then, Integer a is divisible by b if

bMa where M is an integer.

Example

6 30 bydivisibleis

5636 where 5 is an integer.

6 32 bydivisiblenotis

3

16632 where is an integer.

3

16

ExplanationProve, by M.I., is divisible by 8 for all natural numbers n.

132 n

Let P(n) be the proposition

8132 bydivisibleisn

18

1313

,1212

nWhen

P(1) is true

Assume that P(k) is true,

183

8132

2

M

Mk

k

M is an integer.

When n = k+1,

198

898

1189

139

13132

2212

M

M

M

k

kk

9M + 1 is an

integer.

P(k+1) is true.

By the principle of mathematical induction, is divisible by 8 for all natural numbers n.

132 n

Further ExampleProve, by M.I., is divisible by 5 for all natural numbers n.

222 23 nn

Let P(n) be the proposition ‘is divisible by 5. ’

222 23 nn

Show P(1) is true.

,1nwhen

25

10

2323 0221212

Assume that P(k) is true.

Mkk 523 22

where M is an integer.

Consider P(k+1) is true or not.

So hard!!

22

22

2222222

222

22222

21212

295

2559

2429239

4293

23

23

k

k

kkkk

kk

kk

kk

M

M

k is positive

integer.

P(k+1) is true.

1k

By the principle of mathematical induction,

is divisible by 5

for all natural numbers n.

222 23 nn

2229 kM is an integer.