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The Binomial Theorem

The binomial theorem provides a useful method for raising any binomial to a nonnegative integral power. Consider the patterns formed by expanding (x + y)n. (x

+ y)0 = 1 (x + y)1 = x+y
Can you see a pattern? Can you make a guess what the next one would be?

(x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + _x4y + _x3y2 +_ x2y3 + _xy4 + y5
We can easily see the pattern on the x's and the y's. But what about the coefficients? Make a guess and then as we go we'll see how you did.

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

2

Let's list all of the coefficients on the x's and the y's and look for a pattern.

(x + y)5 = 1x5 + 5x4y + 10x3y2 +10 x2y3 + 5xy4 + 1y5 (x + y)0 = 1 (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3

1 1

1

+

1 + 2+ 1

1 +3+ 3+ 1
1 + 4+ 6+ 4+ 1

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
Can you guess the next row?

1

5

10 10

5

1

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

3

Consider the patterns formed by expanding (x + y)n. (x + y)0 = 1 (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3 1 term 2 terms 3 terms

4 terms
5 terms 6 terms

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + x4y + x3y2 +x2y3 + xy4 + y5

Notice that each expansion has n + 1 terms. Example: (x + y)10 will have 10 + 1, or 11 terms.
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

4

Consider the patterns formed by expanding (x + y)n.

(x + y)0 = 1
(x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3 (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

1. The exponents on x decrease from n to 0. The exponents on y increase from 0 to n.
2. Each term is of degree n. Example: The 5th term of (x + y)10 is a term with x6y4.”
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

5

The coefficients of the binomial expansion are called binomial coefficients. The coefficients have symmetry. (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 The first and last coefficients are 1. The coefficients of the second and second to last terms are equal to n. Example: What are the last 2 terms of (x + y)10 ? Since n = 10, the last two terms are 10xy9 + 1y10. The coefficient of xn–ryr in the expansion of (x + y)n is written ? n ? ? ? or nCr . So, the last two terms of (x + y)10 can be expressed ? r ? as 10C9 xy9 + 10C10 y10 or as ?10 ?xy 9 + ?10 ? y10. ? ? ? ?
?9 ? ? ? ?10 ? ? ?
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

6

The triangular arrangement of numbers below is called Pascal’s Triangle. 1 0th row

1 1
1+2=3

1st row

1 2 1 3 3

1 1

2nd row 3rd row

6 + 4 = 10

1 4

6

4

1

4th row

1 5 10 10 5 1

5th row

Each number in the interior of the triangle is the sum of the two numbers immediately above it. The numbers in the nth row of Pascal’s Triangle are the binomial coefficients for (x + y)n .
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

7

Example: Use the fifth row of Pascal’s Triangle to generate the sixth row and find the binomial coefficients ? ?, ? ?
6

? 6? , C ? ? 6 4 ? 1? ? 5?

and 6C2 .

5th row 6th row

1

5

10

10

5

1

1
? 6? ? ? ? 0?
6C0

6
?6? ? ? ?1?
6C1

15
? 6? ? ? ? 2?
6C2

20
? 6? ? ? ? 3?
6C3

15
? 6? ? ? ? 4?
6C4

6
?6? ? ? ?5?
6C5

1
? 6? ? ? ? 6?
6C6

?6? ? ? ?1?

=6=

?6? ? ? ?5?

and 6C4 = 15 = 6C2.

There is symmetry between binomial coefficients. nCr = nCn–r
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8

Example: Use Pascal’s Triangle to expand (2a + b)4. 1 1 1 0th row 1st row

1 2
1 3 1 4 6 3

1
1 4 1

2nd row
3rd row 4th row

(2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4
= 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4 = 16a4 + 32a3b + 24a2b2 + 8ab3 + b4
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

9

The symbol n! (n factorial) denotes the product of the first n positive integers. 0! is defined to be 1.
1! = 1 4! = 4 ? 3 ? 2 ? 1 = 24 6! = 6 ? 5 ? 4 ? 3 ? 2 ? 1 = 720 n! = n(n – 1)(n – 2) ? 3 ? 2 ? 1 Formula for Binomial Coefficients For all nonnegative n! integers n and r,
n

Cr ?
7!

( n ? r )! r !
? 7! 4! ? 3! ?
?

Example: 7 C 3 ?
?

7 4! ? 3!
7 ?6?5?4 4 ? 3 ? 2 ?1
10

( 7 ? 3 )! ? 3!

( 7 ? 6 ? 5 ? 4 ) ? ( 3 ? 2 ? 1) ( 4 ? 3 ? 2 ? 1) ? ( 3 ? 2 ? 1)

? 35

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

Example: Use the formula to calculate the binomial coefficients
10C5, 15C0,

?12 ? and ? 50 ? . ? ? ? ? 48 ? 1? ? ?

10

C5 ?

10! (10 ? 5 )! ? 5! 10! (10 ? 0 )! 0!
?

?

10! 5! ? 5!

?

(10 ? 9 ? 8 ? 7 ? 6 ) ? 5! 5! ? 5!

?

10 ? 9 ? 8 ? 7 ? 6 5 ? 4 ? 3 ? 2 ?1

? 252

10

C0 ?

?

10! 10! 0!
?

?

1! 0!

?

1 1

?1

50! ( 50 ? 49 ) ? 48! 50 ? 49 ? 50 ? 50! ? ? ? ? 1225 ? ?? ? 48 ? ? 48! ? 48! ?1 2! 2! 2 ? ? ( 50 ? 48 )! ? 48!
? 12 ? 12! 12 ? 11! 12 12! ? ? ? ? 12 ? ?? ?1 ? ?1 ?1 1 ? ? (12 ? 1)! ? 1! 1! ! 11! !
Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

11

Binomial Theorem
(x ? y) ? x ? nx
n n n ?1

y ? ? ? nC r x

n?r

y ? ? ? nxy
r

n ?1

? y

n

w ith

n

Cr ?

n! (n ? r )!r !

Example: Use the Binomial Theorem to expand (x4 + 2)3.
( x ? 2 ) ? 3 C 0( x ) ? 3 C1( x ) ( 2 ) ? 3 C 2( x )( 2 ) ? 3 C 3( 2 )
4 3 4 3 4 2 4 2 3

?
? x

( x ) ? 3 x ) ( 2 ) ? 3 x )( 2 ) ? ( 2 ) ( (
4 3 4 2 4 2

3

12

? 6 x ? 12 x ? 8
8 4

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

12

Although the Binomial Theorem is stated for a binomial which is a sum of terms, it can also be used to expand a difference of terms. Simply rewrite (x + y) n as (x + (– y)) n and apply the theorem to this sum.

Example: Use the Binomial Theorem to expand (3x – 4)4.
( 3 x ? 4 ) ? ( 3 x ? ( ? 4 ))
4 4 4

? 1 ( 3 x ) ? 4 ( 3 x ) ( ? 4 ) ? 6 ( 3 x ) ( ? 4 ) ? 4 ( 3 x )( ? 4 ) ? 1 ( ? 4 )
3 2 2 3

4

? 8 1 x ? 4 ( 2 7 x )( ? 4 ) ? 6 ( 9 x )( 1 6 ) ? 4 ( 3 x )( ? 6 4 ) ? 2 5 6
4 3 2

? 81x ? 432 x ? 864 x ? 768 x ? 256
4 3 2

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

13

Example: Use the Binomial Theorem to write the first three terms in the expansion of (2a + b)12 .
? 12 ? ? 12 ? ? 12 ? 12 11 10 2 ? ? ( 2 a ) ? ? ? ( 2 a ) b ? ? ? ( 2 a ) b ? ... ?? ? ?1 ? ?2 ? ?0 ? ? ? ? ?

(2a ? b)

12

?

12! (12 ? 0 )! ? 0!

(2 a ) ?
12 12

12! (12 ? 1)! ? 1!

( 2 a )b ?
11 11

12! (12 ? 2 )! ? 2!

( 2 a ) b ? ...
10 10 2

? (2 a
12

12

) ? 1 2 ( 2 a ) b ? (1 2 ? 1 1 )( 2 a
11 11 10

10

) b ? ...
2

? 4096 a

12

? 2 4 5 7 6 a b ? 1 3 5 1 6 8 a b ? ...
11 10 2

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

14

Example: Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y 0 . The power of y is 1 less than the number of the term in the expansion. The eighth term is 13C7 x 6 y7.
C7 ? 13! 6! ? 7 ! ? ? (13 ? 12 ? 11 ? 10 ? 9 ? 8 ) ? 7! 6! ? 7 ! 13 ? 12 ? 11 ? 10 ? 9 ? 8 6 ? 5 ? 4 ? 3 ? 2 ?1 ? 1716

13

Therefore, the eighth term of (x + y)13 is 1716 x 6 y7.

Copyright ? by Houghton Mifflin Company, Inc. All rights reserved.

15


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