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Quick Way to Expand Binomials Raised to Powers

Date: 11/03/2006 at 01:28:11
From: Daryl
Subject: unique method for binomial expansions

I have a question about a method I once learned to expand a binomial 
raised to a power, i.e., a method to expand something like (x+y)^5 
into a sum of terms with powers of x and y in each term.  

Nearly all textbooks explain that the coefficient for any term of a 
binomial expansion can be found using the equation (n,k) = n!/k!/(n-
k)!, where k corresponds to the power of the second term of the 
binomial.  For (x+y)^5, k corresponds to the power of y in each term 
of the expansion, such that k = 3 for the the third term of the
expansion and the coefficient is equal to (5,3) = 5!/(3!2!) = 10, 
i.e., the third term is equal to 10*x^2*y^3.  

This method using factorials is appropriate when you need to find the
coefficient for a specific term of a binomial expansion, but for
situations when you need to write out the complete expansion term by
term, I use a method taught to me in high school but one I have never
again encountered in any textbook.  For the expansion of (x+y)^5, the
method is applied as follows:

term 1: The coefficient for the first term is equal to 1 and x is 
raised to the fifth power: first term = x^5.

term 2: Find the coefficient for the second term by multiplying the 
coefficient of the first term (=1) by the "x" exponent in the first 
term (=5), then dividing this product by 1 since you are calculating 
using numbers from the first term (term = 1): second term coefficient 
= 1*5/1 = 5.  Decrease the "x" exponent by one and increase the "y" 
exponent by one to complete the expresssion for the second term: 
5*x^4*y.

term 3: Continuing with the same pattern, find the coefficient for 
the third term by multiplying the coefficient of the second term (=5) 
by the "x" exponent in the second term (=4), then dividing this 
product by 2 since you are calculating using numbers from the second 
term (term = 2): third term coefficient = 5*4/2 = 10.  Decrease 
the "x" exponent by one and increase the "y" exponent by one to 
complete the expresssion for the third term: 10*x^3*y^2.

term 4: Find the coefficient for the fourth term by multiplying the 
coefficient of the third term (=10) by the "x" exponent in the third 
term (=3), then dividing this product by 3 since you are calculating 
using numbers from the third term (term = 3): fourth term coefficient 
= 10*3/3 = 10. Decrease the "x" exponent by one and increase the "y" 
exponent by one to complete the expresssion for the fourth term: 
10*x^2*y^3.

term 5: Find the coefficient for the fifth term by multiplying the 
coefficient of the fourth term (=10) by the "x" exponent in the 
fourth term (=2), then dividing this product by 4 since you are 
calculating using numbers from the fourth term (term = 4): fifth term 
coefficient = 10*2/4 = 5. Decrease the "x" exponent by one and 
increase the "y" exponent by one to complete the expresssion for the 
fifth term: 5*x*y^4.

term 6: Find the coefficient for the sixth term by multiplying the 
coefficient of the fifth term (=5) by the "x" exponent in the fifth 
term (=1), then dividing this answer by 5 since you are calculating 
using numbers from the fifth term (term = 5): sixth term coefficient 
= 5*1/5 = 1. Decrease the "x" exponent by one and increase the "y" 
exponent by one to complete the expresssion for the sixth term: 
x^0*y^5 = y^5.  

The many words I've used to explain this method may at first make it 
seem complicated, but it is actually incredibly simple.  I have not 
yet attempted to prove why this method works, but perhaps you already 
are familiar with both the method and the proof?  Also, why do I 
never see this method in any textbooks?



Date: 11/03/2006 at 11:21:05
From: Doctor Greenie
Subject: Re: unique method for binomial expansions

Hi, Daryl --

I personally use this method frequently.  And in my work with high 
school students, both as a substitute teacher and as a math team 
coach, I show it to students every chance I get.

As for why we never see this method taught in textbooks, I'm not 
sure.  It might be because it is not easy to explain in words which 
would "work" in a textbook.  As you found in writing your explanation 
of the method, a good explanation is very wordy--and wordy text is not 
usually very effective in a textbook.

The "proof" follows quite easily from the definition of the C(n,k) 
coefficients.  For example....

                  1
  C(6,0) = 1  (or - )
                  1

           6
  C(6,1) = -
           1

           6*5
  C(6,2) = ---
           1*2

           6*5*4
  C(6,3) = -----
           1*2*3

  ...

It is easy to see that

  C(6,1) = (6/1) * C(6,0)
  C(6,2) = (5/2) * C(6,1)
  C(6,3) = (4/3) * C(6,2)
  ...

And of course, since the rows of Pascal's triangle are symmetric 
(that is, C(n,k) = C(n,n-k)), we only need to use this method to get 
halfway across a row; we can finish the row by writing the preceding 
coefficients in reverse order:

  C(6,0) = 1
  C(6,1) = (6/1)*C(6,0) = 6*1 = 6
  C(6,2) = (5/2)*C(6,1) = (5/2)*6 = 15
  C(6,3) = (4/2)*C(6,2) = (4/3)*15) = 20
  C(6,4) = C(6,2) = 15
  C(6,5) = C(6,1) = 6
  C(6,6) = C(6,0) = 1

The coefficients are

  1 6 15 20 15 6 1

Thanks for sending this to us.  Perhaps I will ask that your question 
and my response get added to the archives so other readers can learn 
this quick and easy way for writing out a row of binomial 
coefficients.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials

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