Asmaa Mahmoud Site Administrator
عدد الرسائل : 982 العمر : 37 Location : egypt / Giza Job/hobbies : math teacher Skills/Courses : thinking Mood : الأوسمة : تاريخ التسجيل : 27/03/2008
| موضوع: Solving a Sequence الإثنين 01 سبتمبر 2008, 11:55 pm | |
| First of all, you should be aware that these problems of determiningformulas for sequences are not well-formed. For example, what's thenext 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 2728 29 30 31 ?You probably think it's 32, but it could be 1. The numbers could bethe days of the year, and after January 31 comes February 1.So you're really looking, in a sense, for the "simplest" formula for thesequence, and "simplest" can be a matter of opinion.In your example, the numbers go up by 6, then 9, then 12, then 15, soI'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 patternbetter). Then, on the line below, list the differences of thosenumbers. On the next line, list the differences of the differences, andso 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 inthe form of a "polynomial".If the first row of differences is all zeroes, then all your numbers arethe same, and the answer is just a constant. The answer looks likethis: A, where "A" is the constant.If the second row of differences is all zeroes, then the answer has theform: 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 "workbackward" to see that the zeroth term would be 0. (The differencebetween 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) ----------- 2Test it:n=0 ==> 0n=1 ==> 3n=2 ==> (6+12)/2 = 9n=3 ==> (9+27)/2 = 18n=4 ==> (12+4/2 = 30n=5 ==> (15+75)/2 = 45and you can try your own numbers.It's too bad that you really need some algebra to solve such problemseasily, but the nice thing is that this method will work for any suchsequence 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 aformula for: 0 + 1 + 2 + 3 + ... + nThe 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 = 0A + B*1 + C*1 = 1A + B*2 + C*4 = 3If you work it out, A = 0, B = 1/2, C = 1/2, so the formula to add upall the numbers from 0 to n is just: (n + n^2) --------- 2 | |
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