Sequence
Arithmetic Sequence
Geometric Sequence
§2.
§3.
§4.
CHAPTER III
SEQUENCE-ARITHMETIC SEQUENCE AND GEOMETRIC SEQUENCE

§1.
Mathematical Induction
Chapter III
SEQUENCE- ARTHMETIC SEQUENCE & GEOMETRIC SEQUENCE
§1.
Mathematical Induction
Review
consider (v)- xét, coi. contain- chứa.
proportion-mệnh đề . Induction-quy nạp.
variable -biến. Deduction-suy diễn.
element- phần tử. Deduce (v) -suy ra
assume/suppose(v)-suy ra. Equality-đẳng thức
relation-mối quan hệ. term-số hạng

New words


nN* : every n belongs to natural numbers, not zero.
xn : x to the (power of) n
x2 : x squared. A>B: A is greater than B, B is less than
y3 : y cubed. M=N: m equals to n
A(n): A of n. 3.x: three times x. 4/x four over x


Activity 1:
For n = 1,2,3,4,5 are P(n), Q(n) true or false
b) nN*, are P(n) , Q (n) true or false
P(n): “
>3n +1 ” và Q(n): “
 n ” với nN*
Consider two proportions containing variables, (variable n)
?
?
Student 1:
Student 2:
P(n) : “ 3n > 3n+1 ”
Q(n) : “ 2n > n ”
Group Activities
Consider two proportions P(n) : “ 3n > 3n+1 ” and Q(n) : “ 2n > n ”
a. For n = 1, 2, 3, 4, 5, are P(n), Q(n) true or false?
b. For every nN*, are P(n), Q(n) false or true ?
Answer:
P(n) : “ 3n > 3n+1 ” Q(n): “ 2n > n ”
b. For every nN* P(n) false; Q(n) we can not sure to say false or true, because we can not check /have not checked for all nN*
3
9
27
81
243
4
7
10
13
16
2
8
16
32
5
4
3
2
1
4
T
T
T
T
T
T
T
T
T
F
For n =1;2;3;4;5
P(n) false
For n =1;2;3;4;5
Q(n) true
Note:
To prove that a proportion is FALSE, we need show at least one case.
To prove that a proportion is TRUE, we need show that it is true for all.
For nN* , when we have checked for many number n (not all), that is not a proof.
Hence, Mathematical Induction , a method of mathematical proof , used to establish a given statement for all natural number , is an important and effective method in mathematics.
§1. MATHEMATICAL INDUCTION
Step 1:
Step 2:
Check to make sure that the proportion is TRUE for n = 1.
Assuming that the statement is TRUE for any natural number n = k  1 (called induction hypothesis ).
I. Mathematical Induction:
Prove that it still holds for n = k + 1. (Using induction hypothesis and all true statements)
Step 3:
Prove that for nN*, then:
1 + 3 + 5 + . . . + (2n – 1) = n2 (1)
Proof:
1) For n = 1, the left side has only one term equal to 1, the right side equals to12.Therefore, (1) is TRUE.
2) Set the left side of (1) equal to Sn. Assume that the equation is true for n = k  1, that is
Sk = 1 + 3 + 5 + . . . + (2k –1) = k2 (induction hypothesis)
3) We have to prove that (1) still holds for n = k+1:
Example1:
II. Examples:
Sk+1=1 + 3 + 5 + …+ (2k – 1) + [2(k + 1) – 1] = (k +1)2
Infact, by the induction hypothesis, we have
Sk+1= Sk+ [2(k + 1) – 1] = k2 + 2k + 1 = ( k + 1)2
Therefore, relation (1) is True for every nN*.
Prove that for every nN*, n3 – n is divisible by 3.
Solution:
Set An = n3 – n (1)
1) For n = 1, then: A1= 0
…
3
2) Suppose that (1) is true for n = k  1, we have:
Ak = (k3 – k)
…
3 (induction hypothesis)
3) We must prove Ak+1
...
3
Ak+1 = (k+1)3- (k+1) = k3 +3k2 +3k +1- k -1
= (k3- k) +3(k2+k)
= Ak+ 3(k2+k)
Ak
…
3 & 3(k2+k)
...
3, hence Ak+1
…
3 .
We conclude that An = n3 – n is divisible by 3 for all nN*.
Ví dụ 2:
Example 2
Prove that for nN*
Group 2:
Prove that for all nN* , then un = 13n –1 6
…
Group 1:
Activity 2: Group Activities
In fact, we have:
For all nN* có un = 13n – 1 6 (2)
…
uk+1 = 13k+1– 1 = 13k .13 –1
= 13k.(12+1) – 1
= 12.13k +13k – 1
= 12.13k + uk
Group 1:
Group 2:
Prove that for every nN*
Solution:
+) For n = 1, then , equality (1) holds.
+) Assume that equality (1) holds for n = k ≥ 1, that is: (Induction hypothesis)
+) Now, we have to prove that (1) is true for n = k + 1, that means :
We rewrite the LHS,
Hence for nN*,
Note:
Exercise 3 (page 82 – Notebook Đại số & Giải tích 11)
Show that for every natural number n  2, we have inequalities : a) 3n > 3n + 1 b) 2n+1 > 2n + 3
Step 1: Check/ show that the proportion is true for n = 2. (not n=1)
Step 2: Suppose that the inequalities holds for natural numbers n = k  2 (induction hypothesis)
Step 3: Prove that the equalities holds for all natural numbers n = k+1.
Closure and Homework

1/ Redo all exercises and examples which have been done in the class. Learning language
2/ Finish the exercises in page 82, 83 notebook.