The Integers Basic Properties of Integers Well-Ordering Principle Every nonempty set S of nonnegative integers contains a least element; that is, there is some integer a in S such that a ≤ b for all b belonging to S. Principle of Mathematical Induction Let S be a set of positive integers with the properties ( i ) 1 belongs to S, and ( ii ) whenever the integer k is in S, then the next integer k + 1 must also be in S. Then S is the set of all positive integers. Sigma Notation Let the Greek letter ∑ indicate the “summation of”, thus we can write the sum of the observations as
n
∑X i =1
i
= x1 + x 2 + .... + x n
The numbers 1 and n are called the lower and upper limits of summation, respectively. Some Results on Summation 1. The summation of the sum of variables is the sum of their summations n
∑ (x i =1
i
n
∑ (a i =1
i
n
n
i =1
i =1
+ y i ) = ∑ xi + ∑ y i n
n
n
i =1
i =1
i −1
+ bi + ... + z i ) = ∑ ai + ∑ bi + ... + ∑ z i
2. If c is a constant, then
n
∑ cx i =1
n
i
= c ∑ xi i =1
3. If c is a constant then n
∑ c = nc i =1
Product Notation The product
x1 x2 x3 ⋅ ⋅ ⋅ xn
n
is
∏x k =1
k
The symbol is read as “ the product of all the xk as k runs from 1 to n.” n
If n is a natural number, then
∏ k =n! k =1
The Binomial Coefficients (x + y)1 = 1x + 1y 2 2 (x + y) = 1x + 2xy + 1y2 (x + y)3 = 1x3 + 3 x2 y + 3xy2 + 1y2 (x + y)4 = 1x4 + 4 x3 y + 6x2y2 + 4xy3 + 1y4
If we copy the coefficients, we obtain what is called Pascal’s triangle: 1 1 1 1
1 2
3 4
1 3
6
1 4
1
…….
Definition: If n is a natural number and k is an integer such that 0≤k≤ n, then we define the binomial coefficient as n! n = k k!(n − k )! Some Properties of Binomial Coefficients: Let n be a natural number. Let k be an integer such that 0 ≤ k ≤ n. Then 1. Symmetry Property n n = k n−k 2. Pascal’s Identity n n n +1 + = if 1 ≤ k ≤ n k −1 k k 3. Binomial Theorem
( x + y)
n
n = ∑ x n−k y k k =0 k n
Note: 1. The first term of the expansion is 2. The second term is
xn .
nx n −1 y .
3. The exponent of x decreases by 1 and the exponent of y increases by 1 as we move from left to right. 4. There are n+1 terms in the expansion. th
5. The n , or the next to last, term of the expansion is 6. The k
th
nxy n −1
term of the expansion (x + y )n is
n n−( k −1) k −1 y x k −1
Example: 1. Use the binomial theorem to obtain the expansion of (2a+b)6. Solution: Let x = 2a and y = b and n = 6. Then 6 6 6 (2a + b)6= (2a)6 + (2a)5 b + (2a)4 b2 0 1 2 6 6 6 + (2a)3 b3 + (2a)2 b4 + (2a) b5 3 4 5 6 + b6. 6 = 64a6 + 6(32a5 )b + 15(16a4 )b2 + 20(8a3 )b3 + 15 (4a2 )b4 + 6 (2a) b5 + b6 = 64a6 + 192a5b + 240a4b2 + 160a3b3 + 60a2b4 + 12ab5 + b6 2. Find the fourth term in the expansion of (2a – b)9. Solution: Given: k = 4 x = 2a y = -b n=9 9 Fourth term: (2a)6(-b)3 = 84(64a6)(-b)3 = -537a6b3 3
Exercises: 1. Write the following in summation notation a. 1 + 2 + 3 + 4 + … + n b. 1 + 3 + 5 + 7 + … + (2n – 1) c. 3 + 5 + 7 + 9 + … + (2n + 1) 2 d. 1 + 4 + 9 + 16 + … + n 2. Write the product of the first n even natural numbers in product notation, and then simplify your result. 3. Write the product of the first n odd natural numbers in product notation, and use factorials to give a formula for this product. 4. Use mathematical induction to prove the following: a. The sum of the first n natural numbers is ½ n(n+1). n n(n + 1) i= ∑ 2 i =1 2 b. 1 + 3 + 5 + … + (2n-1) = n for all n ≥ 1. n
∑ (2i − 1) = n
2
i =1
c.
n
∑i i =1
n
d.
6.
7.
8.
=
n( n + 1)(2n + 1) 6
∑ (2i + 1) = n(n + 2) = n i =1
5.
2
2
+ 2n
Use the binomial theorem to obtain the expansion of the following: 4 a. (2x + 3y) b. (a + b)7. Find the first three terms of the expansion of the following binomials: a. (a + y)33 b. (m –2y)101 Find the indicated power of the number and round off to four decimal places. 5 a. (1 + 0.04) b. 1.054. Find the specified term of the expansion 7 a. fifth term of (x - 2y) 9 b. sixth term of (3x + y) 10 7 c. the term in (x + 2y) that involves x .