1, 7, 23, 55, 109, 191, ___
Date: 10/03/2002 at 10:54:23 From: Luke Wahlstrom Subject: Number pattern My family is stumped on this number pattern. We have tried every way we know to solve it. Could you help us out? 1,7,23,55,109,191,___ (fill in the blank) Thanks! Luke
Date: 10/03/2002 at 15:48:15 From: Doctor Greenie Subject: Re: Number pattern Hi, Luke - Often a sequence like this is created from a formula in the form of a polynomial. If that is the case, the formula can be found by applying a process called the method of finite differences. Here is a link to a page in the Dr. Math archives where there is a detailed discussion of this method: Method of Finite Differences http://mathforum.org/library/drmath/view/53223.html This method is rather advanced for most 11-year-olds; you (or you and your family) are certainly welcome to give it a try if you want to. By just playing around with some ideas, I discovered a way you might be able to find the formula for generating the terms of your sequence without using a method as complicated as the method of finite differences. To be able to do this, you will need the following hint: Compare the given sequence of numbers to the cubes of the first several integers, as indicated in this table: integer n n cubed term in sequence difference ---------------------------------------------------- 1 1 1 0 2 8 7 1 3 27 23 4 4 64 55 9 5 125 109 16 6 216 191 25 ... The differences between n cubed and the n-th term in your sequence show a nice pattern which you can use to develop the formula (or rule) for producing the terms of your sequence. I hope 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/
Date: 10/04/2002 at 11:42:08 From: Luke Wahlstrom Subject: Thank you (Number pattern) Dr Greenie: WE GOT IT! Thank you so much. It was really hard for me and my family. We got an answer of 307. (1,7,23,55,109,191,**307**). You were very helpful. I'm glad we didn't have to use the "advanced" method. Luke
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