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Key equations

Binomial Theorem ( x + y ) n = k 0 n ( n k ) x n k y k
( r + 1 ) t h term of a binomial expansion ( n r ) x n r y r

Key concepts

  • ( n r ) is called a binomial coefficient and is equal to C ( n , r ) . See [link] .
  • The Binomial Theorem allows us to expand binomials without multiplying. See [link] .
  • We can find a given term of a binomial expansion without fully expanding the binomial. See [link] .

Section exercises

Verbal

What is a binomial coefficient, and how it is calculated?

A binomial coefficient is an alternative way of denoting the combination C ( n , r ). It is defined as ( n r ) = C ( n , r ) = n ! r ! ( n r ) ! .

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What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?

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What is the Binomial Theorem and what is its use?

The Binomial Theorem is defined as ( x + y ) n = k = 0 n ( n k ) x n k y k and can be used to expand any binomial.

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When is it an advantage to use the Binomial Theorem? Explain.

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Algebraic

For the following exercises, evaluate the binomial coefficient.

For the following exercises, use the Binomial Theorem to expand each binomial.

( 4 a b ) 3

64 a 3 48 a 2 b + 12 a b 2 b 3

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( 3 a + 2 b ) 3

27 a 3 + 54 a 2 b + 36 a b 2 + 8 b 3

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( 4 x + 2 y ) 5

1024 x 5 + 2560 x 4 y + 2560 x 3 y 2 + 1280 x 2 y 3 + 320 x y 4 + 32 y 5

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( 4 x 3 y ) 5

1024 x 5 3840 x 4 y + 5760 x 3 y 2 4320 x 2 y 3 + 1620 x y 4 243 y 5

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( x 1 + 2 y 1 ) 4

1 x 4 + 8 x 3 y + 24 x 2 y 2 + 32 x y 3 + 16 y 4

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For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.

( a + b ) 17

a 17 + 17 a 16 b + 136 a 15 b 2

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( a 2 b ) 15

a 15 30 a 14 b + 420 a 13 b 2

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( 3 a + b ) 20

3 , 486 , 784 , 401 a 20 + 23 , 245 , 229 , 340 a 19 b + 73 , 609 , 892 , 910 a 18 b 2

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( x 3 y ) 8

x 24 8 x 21 y + 28 x 18 y

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For the following exercises, find the indicated term of each binomial without fully expanding the binomial.

The fourth term of ( 2 x 3 y ) 4

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The fourth term of ( 3 x 2 y ) 5

720 x 2 y 3

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The third term of ( 6 x 3 y ) 7

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The eighth term of ( 7 + 5 y ) 14

220 , 812 , 466 , 875 , 000 y 7

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The seventh term of ( a + b ) 11

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The fifth term of ( x y ) 7

35 x 3 y 4

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The tenth term of ( x 1 ) 12

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The ninth term of ( a 3 b 2 ) 11

1 , 082 , 565 a 3 b 16

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The fourth term of ( x 3 1 2 ) 10

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The eighth term of ( y 2 + 2 x ) 9

1152 y 2 x 7

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Graphical

For the following exercises, use the Binomial Theorem to expand the binomial f ( x ) = ( x + 3 ) 4 . Then find and graph each indicated sum on one set of axes.

Find and graph f 1 ( x ) , such that f 1 ( x ) is the first term of the expansion.

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Find and graph f 2 ( x ) , such that f 2 ( x ) is the sum of the first two terms of the expansion.

f 2 ( x ) = x 4 + 12 x 3

Graph of the function f_2.
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Find and graph f 3 ( x ) , such that f 3 ( x ) is the sum of the first three terms of the expansion.

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Find and graph f 4 ( x ) , such that f 4 ( x ) is the sum of the first four terms of the expansion.

f 4 ( x ) = x 4 + 12 x 3 + 54 x 2 + 108 x

Graph of the function f_4.
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Find and graph f 5 ( x ) , such that f 5 ( x ) is the sum of the first five terms of the expansion.

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Extensions

In the expansion of ( 5 x + 3 y ) n , each term has the form ( n k ) a n k b k ,   where   k successively takes on the value 0 , 1 , 2 , ... , n . If ( n k ) = ( 7 2 ) , what is the corresponding term?

590 , 625 x 5 y 2

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In the expansion of ( a + b ) n , the coefficient of a n k b k is the same as the coefficient of which other term?

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Consider the expansion of ( x + b ) 40 . What is the exponent of b in the k th term?

k 1

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Find ( n k 1 ) + ( n k ) and write the answer as a binomial coefficient in the form ( n k ) . Prove it. Hint: Use the fact that, for any integer p , such that p 1 , p ! = p ( p 1 ) ! .

( n k 1 ) + ( n k ) = ( n + 1 k ) ; Proof:

( n k 1 ) + ( n k ) = n ! k ! ( n k ) ! + n ! ( k 1 ) ! ( n ( k 1 ) ) ! = n ! k ! ( n k ) ! + n ! ( k 1 ) ! ( n k + 1 ) ! = ( n k + 1 ) n ! ( n k + 1 ) k ! ( n k ) ! + k n ! k ( k 1 ) ! ( n k + 1 ) ! = ( n k + 1 ) n ! + k n ! k ! ( n k + 1 ) ! = ( n + 1 ) n ! k ! ( ( n + 1 ) k ) ! = ( n + 1 ) ! k ! ( ( n + 1 ) k ) ! = ( n + 1 k )

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Which expression cannot be expanded using the Binomial Theorem? Explain.

  • ( x 2 2 x + 1 )
  • ( a + 4 a 5 ) 8
  • ( x 3 + 2 y 2 z ) 5
  • ( 3 x 2 2 y 3 ) 12

The expression ( x 3 + 2 y 2 z ) 5 cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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