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| Math Central | Quandaries & Queries |
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Question from Jonathan, a student: I'm trying to find the next number sequence for this equation: 1 11 35 79 149 251, my problem is that I worked it out and ended up with a single number 17. What am I doing wrong. Thank you for any help. |
Jonathan,
Check your arithmetic. I tried successive differences and found the third difference produced a row of sixes.
Penny
If you want to use successive differences, your first difference sequence should be 10, 24, 44, 70, 102 - was it? Then repeating two more times you should (in this case) get a constant third difference.
If you just want to get the next number, extend the constant row and work back up, adding.
If you want a polynomial, constant third differences identify your pattern as cubic. Divide the constant by 3! (=6) to get the cubic coefficient a3.
Then subtract 1a3, 8a3, 27a3,.. from your original numbers to get a quadratic remainder sequence. Repeat again; if the first differences are constant (yours won't be) the quadratic term is 0. Otherwise (if your subtraction is accurate) your second difference will be constant; divide by 2! (=2) to get the quadratic coerfficient a2. And so on.
Once you've identified the pattern as cubic, another way to find coefficients is to solve the linear system
A + B + C + D = 1
8A + 4B + 2C + D = 11
27A + 9B + 3C + D = 35
64A +16B + 4C + D = 79
Good Hunting!
RD
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