26 MULTIPLICATION OF SUMSA proof of the binomial theoremTHE BINOMIAL THEOREM gives the coefficients in the product of n equal binomials: (x + a)^{n} = (x + a)(x + a)· · · (x + a). If we actually multiplied the 4 factors of (x + a)^{4}, then, before adding the like terms, we would find terms in x^{4}, x^{3}a, x^{2}a^{2}, xa^{3}, and a^{4}. The binomial theorem tells how many terms there are of each kind. Those binomial coefficients, the theorem states, are the combinatorial numbers. To prove that, we will first consider the multiplication of any sums; for example, (x + y)(a + b + c). Upon multiplying, we would find six terms. Each term will contain two factors, namely one letter from each factor: xa + xb + xc + ya + yb + yc. Therefore, we can write the product of the following -- (x + y)(a + b)(m + n) -- simply by writing the sum of all combinations of one letter from each factor. xam + xan + xbm + xbn + yam + yan + ybm + ybn. Each term in the product consists of three factors: one from each binomial. Note that there are a total of 2^{3} or 8 terms. In general: Multiplication of n binomials produces 2^{n} terms. For, multiplication of two binomials gives 4 terms: (p + q)(m + n) = pm + pn + qm + qn. If we multiply those with a binomial, we will have 8 terms; those multiplied with a binomial will produce 16 terms; and so on. Example. (x + a)(x + b)(x + c)(x + d)
For, each of the 2^{4} terms will consist of 4 factors: one from each binomial. x^{4} is produced by taking x from each factor. xxxx = x^{4}. There is only one such term. The coefficient of x^{4} is 1. Terms with x^{3} are formed by taking x from any three factors, in every possible way, and the letter from the remaining factor. axxx + xbxx + xxcx + xxxd = (a + b + c + d)x^{3}. The coefficient of x^{3}, therefore, is the sum of the combinations of a, b, c, d taken 1 at a time: a + b + c + d. How many such combinations are there? ^{4}C_{1}: The number of combinations of 4 things taken 1 at a time. Next, terms with x^{2} will come from taking x from two factors, in every possible way, and the letters from the remaining two. There will be ^{4}C_{2} such terms: The number of ways of choosing 2 things -- 2 letters-- from 4. (ab + ac + ad + bc + bd + cd)x^{2}. A term in x will be produced by taking x from 1 factor and the letter from the remaining 3: (abc + abd + acd + bcd)x. There will be ^{4}C_{3} or 4 ways of doing that; of choosing 3 letters from 4. Finally, the constant term will be produced by taking the letter from each of the 4 factors. There is ^{4}C_{4} -- 1 -- way of doing that. The constant term will be abcd. Again: (x + a)(x + b)(x + c)(x + d)
If we now make all the constants equal -- a = b = c = d -- then we have the 4th power of (x + a ):
The binomial coefficients are the number of terms of each kind. In the expansion of (x + a)^{n} with n = 4, they are 1 4 6 4 1. The result is general. The binomial theorem states that in the expansion of (x + a)^{n}, the coefficients are the combinatorial numbers ^{n}C_{k} , where k -- the exponent of a -- successively takes the values 0, 1, 2, . . . , n.
Compare Example 4, Lesson 25. Problem 1. Imagine multiplying out (x + y + z)(a + b + c). a) How many terms would there be? 3^{2} = 9 b) Each term would consist of how many factors? Two Problem 2. Write the product by taking the correct combinations of the integers. a) (x + 1)(x + 3)(x + 4) = x^{3} + (1 + 3 + 4) x^{2} + (1·3 + 1·4 + 3·4)x + 1·3·4 = x^{3} + 8x^{2} + 19x + 12. b) (x − 1)(x + 2)(x − 3) = x^{3} + (−1 + 2 − 3) x^{2} + (−1·2 + −1·−3 + 2·−3)x + (−1·2·−3) = x^{3} − 2x^{2} − 5x + 6. c) (x + 1)(x + 2)(x + 3)(x − 1) = x^{4} + 5x^{3} + 5x^{2} − 5x − 6. Problem 3. In this multiplication (x + 1)(x + 2)(x + 3)(x + 4)(x + 5) what will be the coefficient of x^{3}? 85 Problem 4. (x + a)^{5} = (x + a)(x + a)(x + a)(x + a)(x + a). a) Upon multiplying out, and before collecting like terms, how many
b) How will a term x^{3}a^{2} be produced? By taking a from any two factors, in every possible way, and x from the remaining three factors. c) How many times will that term be produced? In other words, upon
The number of ways of choosing 2 things -- letter a -- from 5: ^{5}C_{2} = 10 Problem 5. In each row of Pascal's triangle, the sum of the binomial coefficients is 2^{n}. Why? 2^{n} is the number of terms upon multiplying n binomials. Each binomial coefficient is the number of terms of that kind. Therefore, the sum of all the terms will be 2^{n}. Next Topic: Mathematical induction Please make a donation to keep TheMathPage online. Copyright © 2021 Lawrence Spector Questions or comments? E-mail: themathpage@yandex.com |