Quick Way to Expand Binomials Raised to PowersDate: 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/ |
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