Sum of a Series
Date: 10/26/96 at 15:18:7 From: Jae Jang Subject: Problem Compute the sum of the coefficients of the expansion of (x+0.5)^100 for which the exponent is divisible by three.
Date: 11/08/96 at 16:07:10 From: Doctor Yiu Subject: Re: Problem Dear Jae, The easiest way to solve this problem is NOT to find this sum all by itself, but rather to find it along with two other numbers. Let's try to write down the expansion of (x+0.5)^100. Instead of writing it in the natural order, let's write: (x+0.5)^n = constant term + the sum of terms of x^3, x^6, x^9, etc up to x^99 PLUS the sum of terms of x, x^4, x^7, etc up to x^100 PLUS the sum of terms of x^2, x^5, x^8, etc up to x^98. All together, there are 101 terms. In the above expression, we collect these terms into THREE classes: (1) Those with exponents divisible by 3. Let's call the SUM of the COEFFICIENTS of these terms A. (2) Those with exponents of the form 3k+1. Let's call the SUM of the COEFFICIENTS of these terms B. (3) Those with exponents of the form 3k+2. Let's call the SUM of the COEFFICIENTS of these terms C. If you substitute 1 for x, then all powers of x are equal to 1 and: A + B + C = 1.5^100 = 3^100/2^100 This is a LINEAR equation in THREE unknowns (A, B, C). Now we want to make two more SUITABLE substitutions to construct two MORE linear equations in A, B, and C. The class (1) of terms are those that look like: 1, x^3, x^6, x^9, up to x^99. If you want to make a substitution to COLLECT the COEFFICIENTS of these terms together, you need: 1 = x^3 = x^6 = x^9 = . . . = x^99. The substitution x=1 above does this precisely. BUT, besides x=1, ANY number x satisfying x^3 = 1 will do. Such an x is called a "cube root of unity." If you are already familar with this term, then you prodeed by considering u and v to be the two COMPLEX cube roots of unity and recall the following useful properties: (i) u^3 = 1 and v^3 = 1 (ii) u = (-1+Sqrt(-3))/2 and v = (-1-Sqrt(-3))/2 (iii) u^2 = v and v^2 = u (iv) 1 + u + v = 0 If you are not familiar with the term "cube root of unity," here is some background info: If x^3 = 1, then x^3 - 1 = 0. Factoring, we have (x-1)(x^2+x+1) = 0. This means x=1, OR, by the quadratic formula: x = (-1+ or - Sqrt(-3))/2 = -0.5 + or - Sqrt(-3)/2. There are two such numbers, and they are complex numbers. I shall call them u (corresponding to the plus sign) and v (corresponding to the minus sign). Note that u = (-1+Sqrt(-3))/2 and v = (-1-Sqrt(-3))/2, which satisfies u^3 = 1 and v^3 = 1. Also, u + 0.5 = Sqrt(-3)/2 and v + 0.5 = -Sqrt(-3)/2. These two numbers u and v have several interesting and useful properties, which can be verified directly, that I listed under (i), (ii), (iii), and (iv) above. If you make the substitution x = u, then (i) All the terms in class (1) have u^3 = u^6 = etc = u^99 = 1. The sum of these terms is equal to A. (ii) All the terms in class (2) have u = u^4 = u^7 = etc = u^100. The sum of these terms is B times u. (iii) All the terms in class (3) have u^2 = u^5 = etc = u^98. The sum of these terms is C times u^2. Since u^2 = v, we write this as C times v. Now, u + 0.5 = Sqrt(-3)/2. This means that: A + Bu + Cv = (Sqrt(-3)/2)^100 = (-3)^50/2^100 = 3^50/2^100. If you make the substitution x = v, then (i) All the terms in class (1) have v^3 = v^6 = etc = v^99 = 1. The sum of these terms is again equal to A. (ii) All the terms in class (2) have v = v^4 = v^7 = etc = v^100. The sum of these terms is B times v. (iii) All the terms in class (3) have v^2 = v^5 = etc = v^98. The sum of these terms is C times v^2. Since v^2 = u, we write this as C times u. Now, v + 0.5 = -Sqrt(-3)/2. This means that: A + Bv + Cu = (-Sqrt(-3)/2)^100 = 3^50/2^100 again. Let's make a summary: A + B + C = 3^100/2^100 A + Bu + Cv = 3^50/2^100 A + Bv + Cu = 3^50/2^100 Of course, you can solve for A, B, C, by systematic elimination. However, property (iv) listed above above helps to get the value of A immediately. Namely, if you add up these three equations, you get: 3A = (3^100+3^50+3^50)/2^100 = (3^100 + 2 times 3^50)/2^100. Dividing by 3, we finally obtain: A = (3^99 + 2 times 3^49)/2^100. This is a precise expression. To get a rough idea of how big this number is, note that the second term (2 times 3^49/2^100) is negligibly small. This means that A is approximately: 3^99/2^100 or 1.3552 times 10^17. -Doctor Yiu, The Math Forum Check out our web site!
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