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1. Mathematical Induction. Case Study. 1.1 The Principle of Mathematical Induction. 1.2Divisibility. Chapter Summary. How can I find the total number of cans in all layers?. That’s easy! We can find a formula to calculate it. Let me explain it to you. - PowerPoint PPT Presentation
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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.
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
Example 1.1TProve, by mathematical induction, that
for all positive integers n.
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
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
∴ The proposition is true for n 1.
Next, assume the proposition is true for some positive integers k, that is,
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
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
Example 1.2T(a) Prove, by mathematical induction, that
for all positive integers n.
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.
∴ The proposition is true for n k 1.
P. 9
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
(a) Prove, by mathematical induction, that
for all positive integers n.
(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.
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
∴ The proposition is true for n 1.
Next, assume the proposition is true for some positive integers k, that is,
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
(a) Prove, by mathematical induction, that
for all positive integers n.
P. 12
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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.
∴ The proposition is true for n k 1.
(a) Prove, by mathematical induction, that
for all positive integers n.
P. 13
1.1 The Principle of1.1 The Principle of Mathematical InductionMathematical Induction
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
(a) Prove, by mathematical induction, that
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
1.2 Divisibility1.2 Divisibility
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.
∴ The proposition is true for n 1.
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
1.2 Divisibility1.2 Divisibility
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
1.2 Divisibility1.2 Divisibility
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.