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Chessboard Squares


Date: 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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