Finding the nth Term
Date: 03/30/2002 at 07:40:43 From: Kathy Angus Subject: Finding the nth term This is the investigation. This was the starting point: A colony of bees is making a honeycomb consisting of hexagonal cells. They start with the center cell. The next day they add the six cells round the center cell, and on the third day they add the remaining ring of cells in the picture. Every day they add a new ring of cells right round the honeycomb. The investigation posed two questions: 1. How many cells will there be at the end of the sixth day? 2. On which day will the number of cells pass the 1000 mark? I managed to find the formula for this. I extended this by using 3D shapes instead of 2D shapes. I used cubes, with day 1 starting with 1 cube; on day 2 I surrounded the first cube with more cubes, both the sides and edges, completely surrounding it. I.e., making a 3x3x3 cube on day 2. I continued this for each day: 3x3x3, 5x5x5, 7x7x7, etc.. I wanted to find the relation between the number of days and the number of NEW ADDITIONAL cubes. These are the results I got: Number of days(n) New additional days(y) 1 0 2 26 3 98 4 218 5 386 The formula I got was: Y = (2n-1)3-(2n-3)3 This works for all except the first day! I would be grateful for any help you give me. Thanks, Kathy
Date: 03/30/2002 at 22:51:05 From: Doctor Peterson Subject: Re: Finding the nth term Hi, Kathy. I would have said that on day 1 you have added 1 block, rather than zero, since you probably had none before beginning. Further, since on each day the total is a complete cube, as on the first and second days, the total on day N is (2N-1)^3, so the number added on day N would be (2N-1)^3 - (2(N-1)-1)^3 which is just what you got. (I see now that you did not get this formula from the numbers, but directly from the problem, which is good!) This can be simplified to [8N^3 - 12N^2 + 6N - 1] - [8N^3 - 36N^2 + 54N - 27] = 24N^2 - 48N + 26 I would check this by making a table including the total each day: day total new formula 0 0^3= 0 - - 1 1^3= 1 1 2 2 3^3= 27 26 26 3 5^3=125 98 98 4 7^3=343 218 218 Now, we can see why day 1 would be anomalous: on the other days, we started with a cube, and added two to each dimension, going to the next odd cube. But on the first day, we didn't start with a cube -1 unit on a side; we started with nothing, and added a single cube. That is, the number on the first day did not arise by following the same rule of surrounding an existing cube, so it is not surprising that your formula doesn't apply. Notice that my formula for the _total_ applies to day 1, but not to my fictional day 0; so that your formula for the number of blocks added applies only _after_ day 1, since it depends on the previous day's total. It is not unusual in mathematics to require a starting condition apart from the formula that applies elsewhere. You just have to say that the formula holds for N > 1. In some cases it makes some sort of sense to continue the pattern back in time before you started, but here that doesn't work, so we don't need to be troubled by the formula not being extensible to N < 2. Incidentally, I don't think your description of the process by which you are building larger blocks is quite right. You are not just adding to the faces and edges, but also to the corners. If you added only cubes that touch on a face, you would have a considerably more interesting problem, since the shapes you are building would not be mere cubes. But the question you raised turns out to be very interesting, and deserves the thought we've given it! It's all too easy to ignore starting conditions and imagine that all formulas work for all cases, and checking against reality is a very good practice to avoid this. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 04/01/2002 at 11:34:09 From: Kathy Angus Subject: Finding the nth term Hello, I would like to thank you very much for the help you gave me with the investigation. It was a great help. I will be using this great site again in the future! Thanks for the quick response! Thanks once again, Kathy
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