the binomial theorem. powers of binomials pascal’s triangle the binomial theorem factorial...

Lesson9.2

The Binomial Theorem

Precalculus

2

2

2

2

3

Use the distributive property to expand the binomial.

1.

2. ( 2 )

3. (2 3 )

4. (2 )

5.

x y

a b

c d

x y

x y

Quick Review

( )( )x y x y 2 2x xy yx y

( 2 )( 2 )a b a b 2 22 2 (2 )a ab ba b 2 24 4a ab b

2 2(2 ) 6 6 (3 )c cd dc d 2 24 12 9c cd d

2 2(2 ) 2 2 ( )x xy yx y 2 24 4x xy y

2( ) ( )x y x y 2 2( 2 )( )x xy y x y 3 2 2 33 3x x y xy y

2 22x xy y

What you’ll learn about

Powers of BinomialsPascal’s TriangleThe Binomial TheoremFactorial Identities

… and whyThe Binomial Theorem is a marvelous study in combinatorial patterns.

Binomial Coefficient

The binomial coefficients that appear

in the expansion of ( )

are the values of for 0,1,2,3,..., .

n

n r

a b

C r n

A classical notation for ,

especially in the context of binomial

coefficients, is .

n rC

n

r

Both notations are read " choose ."n r

Example Using nCr to Expand a Binomial

4

Expand , using a calculator to compute the binomial coefficients.a b

44 3 2 2 3 44 6 4 .a b a a b a b ab b

TI-89 user’s Enter nCr(4,{0,1,2,3,4})

TI-84 user’s Enter 4 nCr {0,1,2,3,4}

This finds the binomial coefficients for n = 4.

The calculator gives the list {1, 4, 6, 4, 1}.

Using these coefficients, construct the expansion:

Pascal’s Triangle

The triangle starts with a pyramid of 1’s.To make a new row, you add 2 entries in the previous row together and put the answer underneath and between the previous entries.You sandwich each row with 1’s.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

When you count rows or entries, start with 0.

Row 0

Row 1

Row 2

Row 3

Row 4

Row 5

0 1 2 3 4 5 Position

If I ask for 5C3, I go to row 5 and choose the 3rd entry

Pascal’s Triangle

How about a Formula

Pascal’s triangle can be used for combinatoric problems. These are problems where you want to see how many different ways you choose a subset from a full group. For example, choosing 3 students from a group of 5.

This is fine for small problems, but isn’t so nice for big ones. How many different ways can I draw 5 cards from a deck of 52? It’s just the 5th entry on the 52nd row of the triangle. OR we can use a formula.

!

! !n m

n nC

m m n m

This isn’t quite the same as a poker hand, since poker only cares about suites when you look at a flush, but you’ll get the idea.

If I want to see how many unique combinations of 5 cards I can make from a deck of 52, then I need 52C5.

52 52! 52!

5 5! 52 5 ! 5!47!

Tip: count the top down to the larger of the 2 numbers in the bottom, then reduce. Only count down the small one in the denominator.

52 51 50 49 48 47!

5 4 3 2 1 47!

The 47! will cancel and you can reduce in the rest of the fraction will reduce like normal. Then just multiply whatever is left.

2,598,960 That’s a lot of different hands!

How about a Formula

Binomial Expansion

Pascal’s triangle can also be used for binomial expansion. The exponent tells you what row to go to and the row entries are the coefficients.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 1

4

4 3 2 2 3 4

4

0

3 2 2 3 4

01 4 6 1

6

4

4 4

x y

x x y x y x y y

x x y x

y

x y

x

y y

The coefficients (blue) come from the triangle.

The red part is never written, but notice how the exponents count down for x and up for y.

Binomial ExpansionThe process

1. Copy the coefficients for the row that matches your exponent.

2. Let the exponent on the first term count down to 0.

3. Let the exponent on the second term count up from 0.

4. If the sign is +, then they are all +5. If the sign is -, then they alternate + and -.

Example Binomial Expansion

5Expand 2 3x y 1. Take the coefficients from row 5.

1 5 10 10 5 12. The (2x) starts with an exponent of 5 and counts down.

1(2x)5 5(2x)4 10(2x)3 10(2x)2 5(2x)1 1(2x)0 3. The (3y) starts with an exponent of 0 and counts up.

1(2x)5(3y)0 5(2x)4(3y)1 10(2x)3(3y)2 10(2x)2(3y)3 5(2x)1(3y)4 1(2x)0(3y)5

4. Since it’s a minus in the middle, we alternate + and – signs.

1(2x)5(3y)0 - 5(2x)4(3y)1 + 10(2x)3(3y)2 - 10(2x)2(3y)3 + 5(2x)1(3y)4 - 1(2x)0(3y)5

5. Now we clean it up. Remember that order of operations says exponents before multiplication.

Example Binomial Expansion (continued)

1(32x5)(1)- 5(16x4)(3y)1 + 10(8x3)(9y2) - 10(4x2)(27y3) + 5(2x)(81y4) - 1(1)(243y5)

5 4 3 2 2 3 4 532 240 720 1080 810 243x x y x y x y xy y

It’s not much fun, but it beats doing this:

52 3

2 3 2 3 2 3 2 3 2 3

x y

x y x y x y x y x y

TI-89 user’sEnter expand((2x – 3y)^5, x)

Be sure your variables are not already assigned a value!!!!

Recursion Formula for Pascal’s Triangle

1 1

1

n n n

r r r

1 1 1n r n r n rC C C

or, equivalently,

Example

8 11Find the coefficient of in the expansion of ( 2) .x x

The only term in the expansion that we need to deal with is

8 311 8 ( 2)C x

3 811!( 2)

8!3!x 3 8165 ( 2) x 81320x

The coefficient is −1320.

The Binomial Theorem

1

For any positive integer ,

... ... ,0 1

!where .

!( )!

nn n n r r n

n r

n

n n n na b a a b a b b

r n

n nC

r r n r

Example

52Expand 2 3x y

5 5 4 3 2 2 3 4 5( ) 5 10 10 5a b a a b a b a b ab b

5We use the Binomial Theorem to expand ( )a b2where a2 3nd .a bx y

5 4 3 22 2

2 2 22 3 4 5

( ) 5( ) ( ) 10( ) ( )

10( ) ( ) 5(

2 2 3 2 3

2 3 2 )( ) ( )3 3

x x y x y

x y x y y

2 52 3( )x y

5 4 2 3 4 2 6 8 1032 240 720 1080 810 243x x y x y x y xy y

Basic Factorial Identities

For any integer 1, n

1 ! 1 !n n n

! 1 !n n n

For any integer 0, n

Homework:

Text pg715 Exercises

# 2-26 even