How many squares in a chessboard n x n

62 + 25 + 16 + 9 + 4 + 1 or (n – 2)(n – 2) + 25 + 16 + 9 + 4 + 1 = 36 + 25 +16 + 9 + 4 + 1, which gives the total number of squares in this particular square. This result will then be added to the next 7 x 7 square.

This is a 7 by 7 square plus previous sequence (36 + 25 +16 + 9 + 4 + 1). It’s mathematical form is 72 + 36 + 25 + 16 + 9 + 4 +1 or (n – 1)(n - ) + 36 + 25 + 16 + 9 + 4 +1 =49 + 36 + 25 + 16 + 9 + 4 +1, which gives the total number of squares in this particular square. This result will then be added to the next 8 x 8 square.

This is 8 x 8 square plus previous sequence (49 +36 + 25 + 16 + 9 + 4 + 1). Since n = 8, as stated earlier, it’s mathematical form is 82 + 49 + 36 + 25 + 16 + 9 + 4 + 1 or

64 + 49 + 36 + 25 + 16 + 9 + 4 +1 = 204.

 

We notice that the 1st square, U1 = 12

                              2nd square, U2 = 12 + 22                             

                              3rd square, U3 = 12 + 22 + 32

                              4th square, U4 = 12 + 22 + 32 + 42                 

                              5th square,U5 =12 + 22 + 32 + 42 + 52

                              6th square, U6 = 12 + 22 + 32 + 42 + 52 + 62

                              7th square, U7 = 12 + 22 + 32 + 42 + 52 + 62 + 72

                              8th square, U8 = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82

From the above sequence, it is seen that the pattern arising is the series of powers of the natural numbers, giving Un = 12 + 22 + ……….+ n2

Therefore the general solution for the sequence above is:

                                       

Un = Un-1 + n2

TESTING THIS SOLUTION

                                                             U4 = U3 + 4          

                                                             U4 = 12 + 22 + 32 + 42

To represent the square sequence of the chessboard in n terms as sum can be mathematically shown as:

                                                    r2 = n/6 [(n+1)(2n+1)]

Where r is any positive integer.

CONDITIONS FOR A CHESSBOARD

1. Drawing a diagonal line (vector line) at 45 degrees from one corner of chessboard to opposite corner of chessboard, gives a line of symmetry diagonally.

 

2. If we fold the chessboard from it’s diagonal line of symmetry, it fits exactly on top of each other.

DIAGRAM(2)

        

3. In a chessboard  of equal sides (square), there are 4 lines of symmetry.

4. Each time you increase the vector line adjoining one square to the next square, it is seen that the square is increased by 1 each time. The square increases from the smallest square, 1x1, then increases to 2x2, 3x3, and so on until nxn. The vector line also cuts through all the corners of the squares as shown below.

   

                           

DIAGRAM(3)

                 

 HOW MANY SQUARES ON A m X n BOARD?

                                               n

                          m

PROBLEM STATEMENT

Using the conditions found for an equal sided chessboard, we will use these conditions to find how many squares there are in a rectangular board, where m and n are not equal sides.

GENERATING DATA AND STRUCTURE

m x n

                                          1 x 2                 Interesting that why are

                                                                                     we left with this part. All will    

                                                  m x n               be revealed in conditions for a       

                           1 x 3               rectangle                     

                         m x n + (n –1)(m –1)

                        2 x 3 +  (3 - 1)(2 - 1)

                           m x n + (n – 1)(m –1)

                           2 x 4  + (4 – 1)(2 – 1)

                           m x n + (n – 1)(m –1)

                           2 x 5  + (5 – 1)(2 – 1)

                                                     m x n + (n – 1)(m – 1)+ (n – 2)(m –2)

                                                     3 x 5  + (5 –1 )(3 – 1) + (5 –2)(3 – 2)

                                                               m x n +  (n – 1)(m –1)+ (n – 2)(m – 2)

                                                        3 x 6 + (6 – 1)(3 – 1) + (6 – 2)(3 – 2)

m = rows, n = columns

Umxn = mxn + (m-1)(n-1) + (m-2)(n-2) +…………..+ (m-m)(n-n)

TESTING FOR A 5 X 4 RECTANGLE

U5X4 = (5 X 4) + (4 X 3) + (3 X 2) + (2 X 1) + (1 X 0)

                        U5X4 = 20 + 12 + 6 + 2 + 0 = 40

If m greater than n

Umxn = (mxn) + (m-1)(n-1) + (m-2)(n-2) + ………+ (m-n)(n-n)

OPENING THE BRACKETS OF THE ABOVE EQUATION

U5X4 = n(mn) – (n-1)(n)(m+n) + (n-1)(n)(2n-1)

                                     2                        6

 

         = n [ mn – (nm + n2 – m – n) + (2n2 – n – 2n + 1)]

                                        2                                  6

         = 6n [ 6mn – 3mn – 3n2 + 3m + 3n + 2n2 – n – 2n + 1]

         = 6n [ 3mn – n2 + 3m + 1]

         = 6n [ 3m( n + 1) – 1(n2 – 1)

         = 6n [ 3m(n + 1) – (n – 1)(n + 1)]

         = 6n (n + 1)(3mn – n + 1)

Therefore, m greater than n

Umxn = 6n(n + 1)(3m – n +1)

If m is less than n

Umxn = (mxn) + (m–1)(n–1)+ (m-2)(n-2)+….+(m-m)(n-m)

OPENINING THE BRACKETS OF THE ABOVE EQUATION

Umxn = m(mn) – m(m – 1) (m + n) + m(m – 1)(2m – 1)

                                         2                            6

     = m [ mn – (m – 1)(m + n) + (m – 1)(2m – 1)]

                                      2                         6

   

         = 6m [ 6mn – 3m2 – 3mn + 3m + 3n + 2m2 – m – 2m +1]

         = 6m [ 3mn – m2 + 3n + 1]

         = 6m [ 3n(m + 1) – 1(m2 – 1)]

         = 6m [ 3n(m + 1) – (m – 1)(m + 1)]

         = 6m (m + 1)(3n – m + 1)

Therefore, for m less than n

Umxn = 6m(m + 1)(3n – m + 1)

 

 

DESCRIPTION OF STRUCTURE OF AN mxn CHESSBOARD

There are two ways of describing the structure of what is happening in a mxn chessboard.

1. Using conditions observed in a nxn chessboard.
2.  To explain the structure of mxn chessboard visually.

   

(1)     CONDITIONS

1. Have to produce a square within the mxn chessboard
2. All corners of the square within the chessboard are align, therefore can draw a 45 degree diagonal line (vector line) cutting through all the corners of the aligned squares from one corner of the chessboard to opposite corner of chessboard.
3. The 45 degree diagonal line cutting through the corners of the squares in the chessboard give a mirror image (when folded diagonally ,lie on top of each other exactly), has a line of symmetry in the diagonal line as well as horizontally and vertically.

Using a 2x3 chessboard to explain the structure and why are we left with (m – 1)(n – 1) and so on.

DIAG (a)                       n=3                                            Umxn = m x n + (m – 1)(n – 1)                            

                  m=2

Using the conditions for a square, the rectangle mxn does not have line of symmetry in the diagonal position. But using the diagonal line (vector line) from the smallest square (1x1) and increasing to the next square (2x2), it fulfils all the conditions mentioned above. However, when we try to make a 3x3 square, we are unable to produce a square because there is no third row to join with the third column. Therefore one column(or row, if nxm) is left over. That’s why in the equation we have –1 in (m – 1)(n – 1). The conditions for m and n are that we are trying to make a square, but there will be a stage when we will unable to produce a square in the mxn rectangle.

If we have a 3x5 rectangle, we can make a squares from 1x1 to 3x3,so we have two

columns left over, which gives us (m – 1)(n – 1) + (m – 2)(m – 2) in the equation.      

DIAG(b)                    n=5

                                                                                  U3X5 = mxn + (m-1)(n-1) + (m-2)(n-2)

                m=3

(2)  VISUAL EXPLANATION

EXAMPLE

Using a 3x5 rectangle to explain the structure.

Diag 1 (a)  Looking at 1x1 squares in the rectangle (mxn)              

                                     n=5  

                 m=3

Diag (b)  Looking at 2x2 squares in the rectangle       Diag (c) Pulling the overlapping squares                      

                                                                                                  apart in the rectangle gives us a

                                                                                                  new value of m and n.

                                                                                                  In this case m and n have

                                                                                                  Decreased by one, (m-1)(n-1).

                 

                      n=5                                                                                      n=4          

                                             

      m=3                                                                                      m=2

                                                                                                       

Diag (d)  Looking at 3x3 squares in the rectangle           Diag (e) Pulling the overlapping squares

                                                                                                      apart in the rectangle shows m  

                                                                                                      and n have decreased by two,

                                                                                                      (m-2)(n-2).

                      n=5                                                                                        n=3

   

         m=3                                                                                      m=1

Therefore, final formula for a 3x5 rectangle chessboard gives

U3x5 = mxn + (m-1)(n-1) + (m-2)(n-2)