Solving a SequenceDate: 12/7/95 at 18:18:24 From: Tina D'Andrea-Conlon Subject: Problem solving with sequences Dear Dr. Math, I am teaching a 7th grade Pre-Algebra class and recently came across the following problem: Write an expression to find the nth term of the following sequence 3, 9, 18, 30, 45 . . . The variable n represents the number of the term, such as 1st, 2nd, 3rd, and so on. My students and I have been able to solve similar problems, but this one has us temporarily stumped. Do you have any hints for us? Thank you, Tina D'Andrea Date: 5/29/96 at 0:53:15 From: Doctor Tom Subject: Re: Problem solving with sequences First of all, you should be aware that these problems of determining formulas for sequences are not well-formed. For example, what's the next number in this sequence: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 ? You probably think it's 32, but it could be 1. The numbers could be the days of the year, and after January 31 comes February 1. So you're really looking, in a sense, for the "simplest" formula for the sequence, and "simplest" can be a matter of opinion. In your example, the numbers go up by 6, then 9, then 12, then 15, so I'll assume the numbers that follow go up by three more each time -- by 18, 21, 24, 27, and so on. I find it easiest to approach such sequences as follows: List your numbers (I'll add a few to your sequence to show the pattern better). Then, on the line below, list the differences of those numbers. On the next line, list the differences of the differences, and so on: n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 3 9 18 30 45 63 84 108 ... 6 9 12 15 18 21 24 ... 3 3 3 3 3 3 ... 0 0 0 0 0 ... If you eventually come to a row of zeroes, you can write the answer in the form of a "polynomial". If the first row of differences is all zeroes, then all your numbers are the same, and the answer is just a constant. The answer looks like this: A, where "A" is the constant. If the second row of differences is all zeroes, then the answer has the form: A + B*n, where "A" and "B" are constants. If the third row of differences is all zeroes, the answer will be: A + B*n + C*n^2, where "A", "B", and "C" are constants. And so on. We just have to figure out what A, B, and C are. In cases like this, it is easier to start with n=0, and we can "work backward" to see that the zeroth term would be 0. (The difference between it and the case where n=1 would be 3.) Now just plug in the first three values: If n=0, A + B*0 + C*0^2 = 0. If n=1, A + B*1 + C*1^2 = 3. If n=2, A + B*2 + C*2^2 = 9. From the first row, A + 0 = 0, so A=0. Using the fact that A = 0, the second row gives: B + C = 3. The third row gives: 2B + 4C = 9. Multiply the equation "B+C=3" by 2 to get: 2B + 2C = 6. Subtract it from "2B+4C=9" to get: 2C = 3. So C = 3/2, and therefore B = 3/2. The equation is: (3n + 3n^2) ----------- 2 Test it: n=0 ==> 0 n=1 ==> 3 n=2 ==> (6+12)/2 = 9 n=3 ==> (9+27)/2 = 18 n=4 ==> (12+48)/2 = 30 n=5 ==> (15+75)/2 = 45 and you can try your own numbers. It's too bad that you really need some algebra to solve such problems easily, but the nice thing is that this method will work for any such sequence problem where some row of differences is all zeroes. As a great example, you might want to try to do the same thing to find a formula for: 0 + 1 + 2 + 3 + ... + n The sequence (and the differences) are just: 0 1 3 6 10 15 21 28 ... 1 2 3 4 5 6 7 ... 1 1 1 1 1 1 ... 0 0 0 0 0 ... A + B*0 + C*0 = 0 A + B*1 + C*1 = 1 A + B*2 + C*4 = 3 If you work it out, A = 0, B = 1/2, C = 1/2, so the formula to add up all the numbers from 0 to n is just: (n + n^2) --------- 2 -Doctor Tom, The Math Forum |
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