Set, Combinatorics, Probability, and Number Theory

Mathematical Structures for

Computer ScienceChapter 3

Copyright © 2006 W.H. Freeman & Co. MSCS Slides Set, Combinatorics, Probability & Number Theory

Section 3.6 Binomial Theorem 2

Pascal’s Triangle

Consider the following expressions for the binomials:(a + b)0 = 1(a + b)1 = a + b(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Row n of the triangle (n 0) consists of all the values C(n, r) for 0 r n. Thus, the Pascal’s looks like this:

Row C(0,0) 0 C(1,0) C(1,1) 1

C(2,0) C(2,1) C(2,2) 2 C(3,0) C(3,1) C(3,2) C(3,3) 3 C(4,0) C(4,1) C(4,2) C(4,3) C(4,4) 4 C(5,0) C(5,1) C(5,2) C(5,3) C(5,4) C(5,5) 5C(n,0) C(n,1) ….………… C(n,n1) C(n,n) n

Section 3.6 Binomial Theorem 3

Pascal’s Triangle

The Pascal’s triangle can be written as: C(0,0)

C(1,0) C(1,1)C(2,0) C(2,1) C(2,2)

C(3,0) C(3,1) C(3,2) C(3,3)

C(4,0) C(4,1) C(4,2) C(4,3) C(4,4)

C(5,0) C(5,1) C(5,2) C(5,3) C(5,4) C(5,5)

. .……….

C(n,0) C(n,1) ………. C(n,n 1) C(n,n)

We note that C(n,k) = C(n 1,k 1) + C(n 1,k) for 1 k n 1

Can be proved simply by expanding the right hand side and simplifying.

Section 3.6 Binomial Theorem 4

Pascal’s Triangle

Coefficients Power

0 0 1 1

1 1 2 1

21 3 3 13

1 4 6 4 1 4

1 5 10 10 5 1 5

Section 3.6 Binomial Theorem 5

Binomial Theorem

The binomial theorem provides us with a formula for the expansion of (a + b)n.

It states that for every nonnegative integer n:

The terms C(n,k) in the above series is called the binomial coefficient.

Binomial theorem can be proved using mathematical induction.

ab n C(n,0)anb0C(n,1)an 1b1C(n,2)an 2b2....C(n,n k)akbn k ....

C(n,n 1)a1bn 1C(n,n)anb0

C(n,k)an kbkk0

n

Section 3.6 Binomial Theorem 6

Exercises

Expand (x + 1)5 using the binomial theoremC(5,0)x5 + C(5,1)x4 + C(5,2)x3 + C(5,3)x4 + C(5,4)x1 + C(5,0)x0

Expand (x 3)4 using the binomial theorem

What is the fifth term of (3a + 2b)7

What is the coefficient of x5y2z2 in the expansion of (x + y + 2z)9 ?

Use binomial theorem to prove that:C(n,0) – C(n,1) + C(n,2) - …….+ (-

1)nC(n,n) = 0