Finding Rules for Number PatternsDate: 06/05/2009 at 22:26:33 From: Jerold Subject: finding the algebraic expression Find the algebraic expression for the following pattern: 3,5,8,12,17,23,30. I found out that it increases by 1 more than before. I tried the different possibilities n^2, 2n, 2^n and trying to add other digits but failed. I have tried similar problems but never found the algebraic expression for this kind yet. Date: 06/06/2009 at 02:35:22 From: Doctor Greenie Subject: Re: finding the algebraic expression Hi, Jerold -- Here is a link to a page in the Dr. Math archives where you can find a discussion of a general method for finding an algebraic expression for a pattern like this: Method of Finite Differences http://mathforum.org/library/drmath/view/53223.html That method is perhaps a bit advanced for a 13-year-old; so let's look at (what I think is) a simpler process that is related to the formal finite differences method. Let's start by looking at the pattern generated by the simple formula t(n) = n^2 For this sequence, we get 1, 4, 9, 16, 25, 36, ... The differences between successive terms are 3, 5, 7, 9, 11, ... In this example, the differences are increasing by 2 each time; and our formula is simply "x^2". Now here is the way you can use this example to find the formula for any similar sequence in which the differences between terms are increasing by the same amount each time: If the differences between successive terms of a sequence are increasing by a constant amount "a" each time, then the formula for generating the sequence is a quadratic polynomial with leading coefficient a/2. I state that rule without proof.... Let's apply it to your example to find the formula for the sequence. The differences in your sequence increase by 1 each time, so the formula for generating the sequence is a quadratic polynomial an^2 + bn + c with leading coefficient 1/2; that is, the formula is of the form (1/2)n^2 + bn + c The "bn + c" is a linear polynomial. If we look at the difference between the value of "(1/2)n^2)" for each value of n and the corresponding term of the given sequence, we can determine that linear polynomial part of our formula. We find n n-th term (1/2)n^2 difference ------------------------------------- 1 3 .5 2.5 2 5 2.0 3.0 3 8 4.5 3.5 4 12 8.0 4.0 5 17 12.5 4.5 ... A bit of examination, or perhaps some formal algebra, shows that the formula in terms of n for the numbers in the difference column above is (1/2)n + 2 And so the complete formula for your sequence is (1/2)n^2 + (1/2)n + 2 It is easy to verify that this formula generates the given sequence; perhaps by using the formula in a slightly different form: (1/2)n^2 + (1/2)n + 2 = (n^2 + n + 4)/2 = (n(n+1)+4)/2 n (n(n+1)+4)/2 ---------------- 1 (1*2+4)/2 = 3 2 (2*3+4)/2 = 5 3 (3*4+4)/2 = 8 4 (4*5+4)/2 = 12 5 (5*6+4)/2 = 17 6 (6*7+4)/2 = 23 7 (7*8+4)/2 = 30 (n^2 + n + 4)/2 - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/