Finding the Rule for a Given SequenceDate: 09/11/2008 at 00:34:25 From: Celina Subject: finding the 20th and 200th term If the terms are -4, 0, 6, 14, 24, 36 what is the rule? Find the 20th and 200th term. Date: 09/11/2008 at 11:45:22 From: Doctor Greenie Subject: Re: finding the 20th and 200th term Hi, Celina -- You don't show any work or indicate what you know about how to solve problems like this, so it's hard to know what to do to help. So let me take a similar sequence and show you how to find the formula ("rule") for the sequence. For a general problem of this type, you can find the formula using the method of finite differences. You can find a detailed description of that method on the following page in the Dr. Math archives: Method of Finite Differences http://mathforum.org/library/drmath/view/53223.html Your example turns out to have a formula that is a quadratic polynomial: t(n) = an^2 + bn + c You can find the formula using the method of finite differences; however, for a sequence defined by a quadratic polynomial, there is an alternative method for finding the formula that requires less work. I will demonstrate the process using the following sequence: 0, 7, 18, 33, 52, ... We start out using the method of finite differences; that is, we write out the terms of the given sequence, then we write a second row of numbers that are the differences between successive terms, then we write out a third row which is the difference between successive numbers in the second row. 0 7 18 33 52 7 11 15 19 4 4 4 The constant "second difference" tells us the formula for the sequence is a polynomial of degree 2. Furthermore--and this is the key to this shortcut method of finding the second degree polynomial formula--the constant second difference is always twice the leading coefficient of the polynomial. In my example, the constant second difference is 4, so we know the formula for the sequence is of the form t(n) = 2n^2 + bn + c The remaining unknown part of the formula is a linear expression. If we rearrange this last formula, we can see a way to determine the rest of the formula by comparing our known "2n^2" to the terms t(n) of the sequence: bn + c = t(n) - 2n^2 So let's evaluate "t(n) - 2n^2" for the first several values of n: n 2n^2 t(n) t(n)-2n^2 ------------------------------ 1 2 0 -2 2 8 7 -1 3 18 18 0 4 32 33 1 5 50 52 2 A quick inspection shows that the formula for the numbers in the last column is n-3 But we know that this is the "bn + c" of our formula; so we know the formula for the sequence is t(n) = 2n^2+n-3 By plugging in the values n=1 to n=5, you can easily verify that this formula produces the original sequence of numbers. Now see if you can use this method for finding the formula for your sequence.... Note that in my example the expression for the n-th term is factorable: 2n^2+n-3 = (2n+3)(n-1) If this is the case, then sometimes we can discover the formula for the sequence by looking for a pattern in the possible factorizations of the given numbers. For my example sequence, we have 0 = (anything)*0 7 = 1*7 18 = 2*9 or 3*6 33 = 3*11 52 = 2*26 or 4*13 If you examine these possible factorizations (ignoring the first "0" term for the moment), you can find the pattern 1*7; 2*9; 3*11; 4*13 and then for the initial "0" term, to keep the pattern intact, you can write the first 5 terms of the sequence as 0*5; 1*7; 2*9; 3*11; 4*13 Then, by examining the patterns for each of the two factors for each term, we can find 0, 1, 2, 3, 4, ... --> (n-1) 5, 7, 9, 11, 13, ... --> (2n+3) And so we can determine the formula for the sequence to be t(n) = (n-1)(2n+3) = 2n^2+n-3 It turns out your sequence is also defined by a quadratic polynomial that is factorable--so you might also be able to find the formula for your example by this factorization process. I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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