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  • Level: GCSE
  • Subject: Maths
  • Word count: 1890

How many squares in a chessboard n x n

Extracts from this document...

Introduction

HOW MANY SQUARES ON A CHESSBOARD (n x n)?

                                                n

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PROBLEM STATEMENT

We were asked to find how many squares were on a chessboard. By observation a chessboard has 8 rows by 8 columns of squares plus one big square, which is the board itself. This will give 65 squares as the answer. However looking at the question in depth, it is seen that many squares can be formed within in the chessboard. The most systematic approach to find exactly  how many squares there are in a chessboard, would be to start from the smallest square to the largest square. We know that a square has 4 equal sides, therefore we can move up n row(s) and across n column(s) on the chessboard, which gives us (n x n) and is also equal to n2 (n squared).

Since the chessboard is an 8 by 8, we can rewrite (n x n) as 8 x 8, where n = 8. So for a single square we write (n – 7)(n –7), where n is 8 in the equation, giving (8 – 7)(8 –7) =1 square.

GENERATING DATA AND DESCRIPTION OF STRUCTURE

Applying the above information, we decided to generate the data starting from the smallest square, which is a 1 x 1 square to the largest square, which is n x n.

This is a one by one square and can be represented mathematically as 12 (1 squared) or

(n – 7)(n – 7) =1 x 1 = 1,       where n = 8

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...read more.

Middle

(n – 4)(n – 4) + 9 + 4 + 1 = 16 + 9 + 4 + 1, which gives the total number of squares in this particular square. This result will be added to the next 5 x 5 square.

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This is 5 by 5 square plus previous sequence (16 + 9 + 4 + 1). It’s mathematical form is

52 + 16+ 9 + 4 + 1 or (n – 3)(n – 3) + 16 +9 + 4 + 1 = 25 +16 + 9 + 4 +1, which gives the total number of squares in this particular square. This result will be added to the next 6 x 6 square.

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This is 6 by 6 square plus previous sequence (25 + 16 + 9 + 4 + 1). It’s mathematical form is

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

...read more.

Conclusion

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)

...read more.

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