Finding a Formula for a Number Pattern
Date: 09/30/2004 at 00:14:28 From: Kelly Subject: Sequence pattern/factoring Hi Dr. Math, I am in a geometry class and we are learning about sequences and how to find the patterns in numbers. I was fine up until our teacher wrote this up on the board: 0, 3, 8, 15, 24, 35 She told us that we had to use factoring to find this answer. What is confusing about this question is I don't know how to get from the problem to the answer that was given: (n+1)(n-1) How did my teacher get this? I know from this problem that the whole pattern is that you add by odd numbers (like to get from 0 to 3 you go by 3, to get from 3 to 8 you go by 5, 8 to 15 you go by 7, etc). But I can't see how to get that other answer. Any help would be greatly appreciated. Thanks for your time.
Date: 09/30/2004 at 09:05:23 From: Doctor Peterson Subject: Re: Sequence pattern/factoring Hi, Kelly. The hint says to factor the numbers in the sequence: 0, 3, 8, 15, 24, 35 0, 3, 2*2*2, 3*5, 2*2*2*3, 5*7 What if we look for a way to factor each number into TWO factors so that they all look similar? Here is what I come up with: 0, 1*3, 2*4, 3*5, 4*6, 5*7 When I saw this, I just noticed that the pattern was one of multiplying a pair of numbers where both increase by one each time, with the first number going 1,2,3... and the second going 3,4,5..., etc.. One helpful way to organize your thoughts is to make a chart of the pattern and include the index, or the number of the result in the pattern. Then you can see if there is a consistent relationship between the pattern and the index. In this case, you can see that the first number is always one less than the index, and the second is one more: index term ----- ---- 1 0*2 2 1*3 3 2*4 4 3*5 5 4*6 6 5*7 n (n-1)*(n+1) The second term is 2-1 times 2+1, the third is 3-1 times 3+1, and so on. Seeing that this pattern applies to all the terms, the formula for the sequence is a[n] = (n-1)(n+1) (Of course, any answer is really just a guess--this is a nice formula, so we assume it is what was intended, but in real life sequences aren't always nice!) How did the teacher get it? Probably because she or whoever invented the problem STARTED with (n-1)(n+1) and wrote out that sequence! Solving problems like this is very different from making them up. You couldn't just write down any random sequence and then figure out "the pattern" to it; random numbers don't generally have any pattern. The pattern is there because someone deliberately made it that way. Now, you seem to have come up with a perfectly good alternative answer, using a standard technique of looking for a way to get from one term to the next. The sequence of differences is 0, 3, 8, 15, 24, 35 3, 5, 7, 9, 11 and that pattern is clear enough that you know how to find as many terms as you want. That's a fine answer, unless you were asked to find an explicit formula, which means you can find any term in the sequence by calculating it directly and not having to work your way through the sequence all the way to the desired term. How could you work your pattern into an explicit formula? I happen to know that the sum of consecutive odd numbers is a square: 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 and so on. The formula for this is 1 + 3 + 5 + ... + (2n-1) = n^2 Your sequence is just my sum with 1 removed from the start! So the formula is a[n] = n^2 - 1 Check that: if we add 1 to each term of your sequence, we get 0, 3, 8, 15, 24, 35 1, 4, 9, 16, 25, 36 (In fact, that is another way one could find the answer!) And this new formula is equivalent to the one you were told, since the difference of squares n^2 - 1 factors as (n-1)(n+1). Interesting, isn't it? There are a lot of ways to solve this one, and they all give the same answer. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 09/30/2004 at 21:03:50 From: Kelly Subject: Sequence pattern/factoring Hi Dr. Math, Thanks for answering my question, and the second idea of how this problem can be viewed. It is really interesting to see what you can do with some problems like these.
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