26 ## MULTIPLICATION OF SUMSTHE BINOMIAL THEOREM gives the coefficients in the product of ( If we expanded ( then before adding the like terms, we would find terms in
What the binomial theorem does is tell We will see that determining those coefficients depends on the theory of combinations. First, then, we will consider the multiplication of any sums; for example, ( The final product will have this form:
As with the binomial theorem, the question is: What are the coefficients? They will be some function of the constants 1, 2, 3. For the moment, we will simply show the result:
The coefficient of The coefficient of And the constant term is their combination taken Why those combinations? We will see why as we continue. For the moment, the student should attempt this problem.
Problem 1. Multiply out ( To see the answer, pass your mouse over the colored area.
Let us now begin again, and analyze the multiplication of these elementary sums: ( When we multiply out, we will find six terms. Each term will contain two factors, namely one letter from each factor:
Therefore, to multiply the following, ( simply write the sum of all combinations of one letter from each factor:
Each term in the product consists of three factors: one from each binomial. Now consider these four binomial factors: ( Let us see how each term in the product will be produced. If we were actually to multiply out, we would find 2 ( If we multiply those with a binomial, we will have 8 terms; and finally those multiplied with a binomial will produce 16 terms. In general: Multiplication of n binomials
produces 2 Now, in the multiplication of those four binomials, each of those 2 It remains only to determine the coefficients. ( How then is a term Terms with
+ xbxx
+ xxcx
= (xxxda + b + c + d)x^{3}The coefficient of Terms with (
= ( The coefficient of The coefficient of
abc
+ abd
+ acd
+ bcd)xFinally, the term independent of We have, (
Notice again that each term has four factors. Example. In this multiplication, ( what number will be the coefficient of
The coefficient of The binomial coefficients Now consider the product of these four equal binomials: ( That is, let us expand ( Again, if we were to actually multiply out, then
and so on. That is, there will be terms in The binomial coefficients are Now, how is a term How will a term ( By taking
Therefore the number of terms equal to A term
Therefore, the number of terms equal to A term And finally, a term This explains the binomial coefficients for the expansion of ( Again, the binomial coefficients are how many terms there are of each kind. They are none other than the
combinatorial numbers This result is general. The binomial theorem states that in the expansion of
(
Problem 2. Imagine multiplying out ( a) How many terms would there be?
3 b) Each term would consist of how many factors? Two
Problem 3. Imagine multiplying out ( a) How many terms would there be?
2 Thus a product of b) Each term would consist of how many factors? Four c) How is a term produced that contains three factors of
By taking d) Therefore, what is the coefficient of e) What is the coefficient of f) What is the coefficient of g) What is the coefficient of h) What is the constant term?
Problem 4. Multiply out by taking the correct combinations of the integers. a) ( b) (
Problem 5. In this multiplication (
Problem 6. ( a) Upon multiplying out, and before collecting like terms, how many
b) How will a term
By taking c) How many
Problem 7. In each row of Pascal's triangle, the sum of the binomial coefficients is 2
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