Chessboard SquaresDate: Mon, 7 Nov 1994 13:02:59 -0800 (PST) From: Barbra Moore Dear Dr. Math, We are students from Sherwood Elementary in Edmonds, WA. We have been working on problems in which we investigate patterns and functions. We worked on a problem called the chessboard squares. We discovered that there are 204 squares on the board and we found several ways to look at it. We found that you would add the different squares - 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64. We did this and it worked because of the pattern, but our teacher wants to know why it works and if there is a quick way to get to the answer. What if there was a 100 by 100 board? Is there a way to use a function to find this out? We would love to hear from you. 6th grade students in Mrs. Moore's room. From: Dr. Ken Subject: Re: Chessboard squares Date: Tue, 8 Nov 1994 13:05:24 -0500 (EST) Hello there! I think I figured out what you mean, and I think it's a fantastic question! You mean the number of squares of any size, right? So in the above sum, the 1 corresponds to the number of 8x8 squares, the 4 is the number of 7x7 squares, and so on, right? Let us know if it's not. Here's a clue about why it works. Picture the chessboard tacked up on a wall. Then let's say you have a 5x5 square sitting in the lower left corner. Where else can you slide that square to? Well, you could keep it there, or you could slide it one space to the right, or two spaces to the right, or three spaces to the right; or you could slide it up zero, up one, up two, or up three spaces. Also, you could do both, and slide it up one and over two, or any of the various combinations of slidings. So there are four options for sliding in each of the two directions, right? Zero spaces, one spaces, two spaces, or three. So the number of different places you could put that 5 by 5 square is 4 times 4 (4 in each of the two directions). How many different places could you put a 4 by 4 square on the board? Can you come up with a formula, so that if I tell you the size of the square (like five by five, or four by four), you could tell me the number of different places you could put it in the chessboard? The second part of your question is also very interesting. I'm very impressed that you're doing it in elementary school; I didn't see that problem until high school! You want to know a formula for the sum of the first n perfect squares. Well, it's kind of tricky to derive, but I do happen to know what it is. Ordinarily, I wouldn't just give it to you without some kind of proof or derivation, or at least some hint of what's actually going on, but the only proof I know is, I think, too sophisticated for elementary school (If I'm wrong, PLEASE let us know! I don't want to underestimate your capabilities!). The proof uses a concept called mathematical induction. Anyway, the formula for the sum of the first n perfect squares is n x (n + 1) x (2n + 1) ______________________ 6 Check it out to make sure that it works. If you have any more questions, send them on in! -Ken "Dr." Math From: Jason Date: Sat, 13 Sep 1997 22:36:07 -0400 Subject: Doctor Math Hi, You said the formuala was n(n+1)(2n+1) ------------- 6 which works out perfectly. I have some questions though. Where is your work, how did you get that answer, how did you get divided by six, how did you get your proof? Thank you Jason Subject: Re: Doctor Math Date: Mon, 15 Sep 1997 17:26:26 -0400 (EDT) Hi Jason - Here are two things that might help you understand this problem. The first is a proof of why the formula works. The second is something that might help your intuitive sense of why it works. I wish I knew a nice way to explain the formula, something like "the n(n+1) comes from the triangular numbers, and then you multiply by (2n+1)/6 because...." Unfortunately I don't know how to explain the formula like that. Here's the proof, using a concept called "mathematical induction": Assume that the formula works for the first few values of n (which you can check just by plugging in numbers). We'll show that if it works for n, then it works for n+1. So if it works for n=5, it works for n=6. And if it works for n=6, it works for n=7. And so on. This will show that the formula works no matter how high we go, so it works for all values of n. Here's how it works: if the sum of the first n squares is n(n+1)(2n+1)/6, then let's add another square. So the sum of the first n+1 squares is: n(n+1)(2n+1) 6(n+1)^2 + n(n+1)(2n+1) ------------ + (n+1)^2 = ----------------------- 6 6 (n+1)[6(n+1) + n(2n+1)] Factor out (n+1): = ------------------------- 6 (n+1)[6n + 6 + 2n^2 + n] = ------------------------- 6 (n+1)[2n^2 + 7n + 6] = -------------------- 6 (n+1)[(n+2)(2n+3)] Factor: = ------------------- 6 (n+1)(n+1 + 1) (2(n+1) + 1) = ------------------------- 6 Now notice that this is what we would get if we plugged n+1 in for n in the original formula. So the formula works for n+1, and we're done. Mathematical induction is very powerful, but it's sometimes hard to get the hang of. If you didn't understand all of that or why it works, maybe you can get a math teacher to explain it in a different way. The other part is more intuitive: picture adding up your squares as building a pyramid out of blocks, each block 1 unit on a side. The tip of the pyramid has 1 block, the next level down has 4 blocks (a 2x2 square), the next level has 9 blocks (a 3x3 square), and so on. To find the total number of blocks, we could try to measure the volume of the pyramid, and that would give us a pretty good estimate. The taller we build our pyramid, the better this estimate will be (do you see why?). So we can use the volume of a pyramid: base*height/3. The base has area n^2, the height is n, so the volume is n^3/3. What about our formula? As n gets bigger and bigger, n+1 starts to look a lot like n, and 2n+1 starts to look a lot like 2n. A difference of 1 in large numbers is a lot less important than a difference of 1 in small numbers. So our formula starts to look like this as n gets big: n(n)(2n) -------- 6 And that simplifies to n^3/3. |
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