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Binomial Theorem |
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A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find (x+y) n
without multiplying the binomial by itself n
times.
In Counting Principles{: .target-chapter}, we studied combinations{: data-type="term" .no-emphasis}. In the shortcut to finding (x+y) n ,
we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation ( n r )
instead of C(n,r),
but it can be calculated in the same way. So
The combination ( n r )
is called a binomial coefficient. An example of a binomial coefficient is ( 5 2 )=C(5,2)=10.
and r
are integers greater than or equal to 0 with n≥r,
then the binomial coefficient is
Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.
-
( 5 3 )
-
( 9 2 )
-
( 9 7 ) {: type="a"}
function on your calculator.
-
( 9 2 )= 9! 2!(9−2)! = 9⋅8⋅7! 2!7! =36
-
( 9 7 )= 9! 7!(9−7)! = 9⋅8⋅7! 7!2! =36 {: type="a"}
- ( 11 4 ) {: type="a"}
When we expand (x+y) n
by multiplying, the result is called a binomial expansion{: data-type="term"}, and it includes binomial coefficients. If we wanted to expand (x+y) 52 ,
we might multiply (x+y)
by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.
First, let’s examine the exponents. With each successive term, the exponent for x
decreases and the exponent for y
increases. The sum of the two exponents is n
for each term.
Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:
These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.
Another way to see the coefficients is to examine the expansion of a binomial in general form, x+y,
to successive powers 1, 2, 3, and 4.
Can you guess the next expansion for the binomial (x+y) 5 ?
{: #CNX_Precalc_Figure_11_06_002}
See [link], which illustrates the following:
-
There are n+1
terms in the expansion of (x+y) n .
-
The degree (or sum of the exponents) for each term is n.
-
The powers on x
begin with n
and decrease to 0.
-
The powers on y
begin with 0 and increase to n.
-
The coefficients are symmetric.
To determine the expansion on (x+y) 5 ,
we see n=5,
thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of x,
the pattern is as follows:
-
Introduce x 5 ,
and then for each successive term reduce the exponent on x
by 1 until x 0 =1
is reached.
-
Introduce y 0 =1,
and then increase the exponent on y
by 1 until y 5
is reached.
x 5 , x 4 y, x 3 y 2 , x 2 y 3 , x y 4 , y 5
The next expansion would be
But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle{: data-type="term" .no-emphasis}, shown in [link].
{: #CNX_Precalc_Figure_11_06_001}
To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add 1+1
to find the middle number, 2. In the nth
row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.
To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.
-
Determine the value of n
according to the exponent.
-
Evaluate the k=0
through k=n
using the Binomial Theorem formula.
-
Simplify. {: type="1"}
-
(x+y) 5
-
( 3x−y ) 4 {: type="a"}
into the formula. Evaluate the
<math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> <mi>k</mi><mo>=</mo><mn>0</mn> </mrow> </math>
through
<math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> <mi>k</mi><mo>=</mo><mn>5</mn> </mrow> </math>
terms. Simplify.
<div data-type="equation" id="eip-id1165135600195" class="unnumbered" data-label="">
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"> <mrow> <mtable columnalign="left"> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <msup> <mrow> <mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mi>y</mi><mo stretchy="false">)</mo> </mrow> <mn>5</mn> </msup> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mo>=</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>5</mn> </msup> <msup> <mi>y</mi> <mn>0</mn> </msup> <mo>+</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>4</mn> </msup> <msup> <mi>y</mi> <mn>1</mn> </msup> <mo>+</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>3</mn> </msup> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>+</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>3</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>2</mn> </msup> <msup> <mi>y</mi> <mn>3</mn> </msup> <mo>+</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>4</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>1</mn> </msup> <msup> <mi>y</mi> <mn>4</mn> </msup> <mo>+</mo><mrow><mo>(</mo> <mrow> <mtable> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> <mtr> <mtd> <mn>5</mn> </mtd> </mtr> </mtable> </mrow> <mo>)</mo></mrow><msup> <mi>x</mi> <mn>0</mn> </msup> <msup> <mi>y</mi> <mn>5</mn> </msup> </mrow> </mtd> </mtr> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <msup> <mrow> <mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mi>y</mi><mo stretchy="false">)</mo> </mrow> <mn>5</mn> </msup> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mo>=</mo><msup> <mi>x</mi> <mn>5</mn> </msup> <mo>+</mo><mn>5</mn><msup> <mi>x</mi> <mn>4</mn> </msup> <mi>y</mi><mo>+</mo><mn>10</mn><msup> <mi>x</mi> <mn>3</mn> </msup> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>+</mo><mn>10</mn><msup> <mi>x</mi> <mn>2</mn> </msup> <msup> <mi>y</mi> <mn>3</mn> </msup> <mo>+</mo><mn>5</mn><mi>x</mi><msup> <mi>y</mi> <mn>4</mn> </msup> <mo>+</mo><msup> <mi>y</mi> <mn>5</mn> </msup> </mrow> </mtd> </mtr> </mtable> </mrow> </math>
</div>
-
Substitute n=4
into the formula. Evaluate the k=0
through k=4
terms. Notice that 3x
is in the place that was occupied by x
and that –y
is in the place that was occupied by y.
So we substitute them. Simplify.
(3x−y) 4 =( 4 0 ) (3x) 4 (−y) 0 +( 4 1 ) (3x) 3 (−y) 1 +( 4 2 ) (3x) 2 (−y) 2 +( 4 3 ) (3x) 1 (−y) 3 +( 4 4 ) (3x) 0 (−y) 4 (3x−y) 4 =81 x 4 −108 x 3 y+54 x 2 y 2 −12x y 3 + y 4
{: type="a"}
raised to odd powers is negative, but (−y)
raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.
-
(x−y) 5
-
(2x+5y) 3 {: type="a"}
- 8 x 3 +60 x 2 y+150x y 2 +125 y 3 {: type="a"}
Expanding a binomial with a high exponent such as (x+2y) 16
can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.
Note the pattern of coefficients in the expansion of (x+y) 5 .
The second term is ( 5 1 ) x 4 y.
The third term is ( 5 2 ) x 3 y 2 .
We can generalize this result.
term of the binomial expansion{: data-type="term" .no-emphasis} of (x+y) n
is:
-
Determine the value of n
according to the exponent.
-
Determine (r+1).
-
Determine r.
-
Replace r
in the formula for the (r+1)th
term of the binomial expansion. {: type="1"}
without fully expanding the binomial.
we will use r=9
in our calculations.
without fully expanding the binomial.
- The Binomial Theorem
- Binomial Theorem Example {: data-display="block"}
| Binomial Theorem | (x+y) n = ∑ k−0 n ( n k ) x n−k y k
| | (r+1)th
term of a binomial expansion | ( n r ) x n−r y r
| {: #eip-id1165134166609 summary=".."}
-
( 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].
It is defined as ( n r )= C(n,r) = n! r!(n−r)! .
and can be used to expand any binomial.
For the following exercises, evaluate the binomial coefficient.
For the following exercises, use the Binomial Theorem to expand each binomial.
For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.
For the following exercises, find the indicated term of each binomial without fully expanding the binomial.
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.
such that f 1 (x)
is the first term of the expansion.
such that f 2 (x)
is the sum of the first two terms of the expansion.
such that f 3 (x)
is the sum of the first three terms of the expansion.
such that f 4 (x)
is the sum of the first four terms of the expansion.
such that f 5 (x)
is the sum of the first five terms of the expansion.
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?
the coefficient of a n−k b k
is the same as the coefficient of which other term?
What is the exponent of b
in the kth
term?
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)!.
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)! + kn! k( k−1 )!( n−k+1 )! = ( n−k+1 )n!+kn! k!( n−k+1 )! = ( n+1 )n! k!( ( n+1 )−k )! = ( n+1 )! k!( ( n+1 )−k )! =( n+1 k )
-
( x 2 −2x+1)
-
( a +4 a −5) 8
-
( x 3 +2 y 2 −z) 5
-
(3 x 2 − 2 y 3 ) 12
cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.
binomial coefficient : the number of ways to choose r objects from n objects where order does not matter; equivalent to C(n,r),
denoted ( n r ) ^
binomial expansion : the result of expanding (x+y) n
by multiplying ^
Binomial Theorem : a formula that can be used to expand any binomial