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  • 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

( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5 .

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 , shown in [link] .

Pascal's Triangle

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3 rd row, flank the ends of the rows with 1’s, and add 1 + 1 to find the middle number, 2. In the n th 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.

Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents

The binomial theorem

The Binomial Theorem    is a formula that can be used to expand any binomial.

( x + y ) n = k = 0 n ( n k ) x n k y k = x n + ( n 1 ) x n 1 y + ( n 2 ) x n 2 y 2 + ... + ( n n 1 ) x y n 1 + y n

Given a binomial, write it in expanded form.

  1. Determine the value of n according to the exponent.
  2. Evaluate the k = 0 through k = n using the Binomial Theorem formula.
  3. Simplify.

Expanding a binomial

Write in expanded form.

  1. ( x + y ) 5
  2. ( 3 x y ) 4
  1. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify.
    ( x + y ) 5 = ( 5 0 ) x 5 y 0 + ( 5 1 ) x 4 y 1 + ( 5 2 ) x 3 y 2 + ( 5 3 ) x 2 y 3 + ( 5 4 ) x 1 y 4 + ( 5 5 ) x 0 y 5 ( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5
  2. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3 x 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.
    ( 3 x y ) 4 = ( 4 0 ) ( 3 x ) 4 ( y ) 0 + ( 4 1 ) ( 3 x ) 3 ( y ) 1 + ( 4 2 ) ( 3 x ) 2 ( y ) 2 + ( 4 3 ) ( 3 x ) 1 ( y ) 3 + ( 4 4 ) ( 3 x ) 0 ( y ) 4 ( 3 x y ) 4 = 81 x 4 108 x 3 y + 54 x 2 y 2 12 x y 3 + y 4
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Write in expanded form.

  1. ( x y ) 5
  2. ( 2 x + 5 y ) 3
  1. x 5 5 x 4 y + 10 x 3 y 2 10 x 2 y 3 + 5 x y 4 y 5
  2. 8 x 3 + 60 x 2 y + 150 x y 2 + 125 y 3
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Using the binomial theorem to find a single term

Expanding a binomial with a high exponent such as ( x + 2 y ) 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 .

( x + y ) 5 = x 5 + ( 5 1 ) x 4 y + ( 5 2 ) x 3 y 2 + ( 5 3 ) x 2 y 3 + ( 5 4 ) x y 4 + 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.

( n r ) x n r y r

The (r+1)th term of a binomial expansion

The ( r + 1 ) th term of the binomial expansion    of ( x + y ) n is:

( n r ) x n r y r

Given a binomial, write a specific term without fully expanding.

  1. Determine the value of n according to the exponent.
  2. Determine ( r + 1 ) .
  3. Determine r .
  4. Replace r in the formula for the ( r + 1 ) th term of the binomial expansion.

Writing a given term of a binomial expansion

Find the tenth term of ( x + 2 y ) 16 without fully expanding the binomial.

Because we are looking for the tenth term, r + 1 = 10 , we will use r = 9 in our calculations.

( n r ) x n r y r
( 16 9 ) x 16 9 ( 2 y ) 9 = 5 , 857 , 280 x 7 y 9
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Find the sixth term of ( 3 x y ) 9 without fully expanding the binomial.

10 , 206 x 4 y 5

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Access these online resources for additional instruction and practice with binomial expansion.

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
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Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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salma
Commplementary angles
Idrissa Reply
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Nharnhar
Practice Key Terms 3

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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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