• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  • 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

DIAGRAM 1image01.pngimage01.pngimage01.pngimage00.pngimage00.png

image00.png

image01.pngimage00.pngimage01.pngimage01.pngimage01.png

image00.pngimage00.pngimage00.pngimage01.png

image02.pngimage01.pngimage01.pngimage00.png

image01.pngimage00.pngimage01.pngimage00.pngimage01.png

image00.pngimage01.pngimage01.png

image03.png

image04.pngimage00.pngimage01.pngimage00.pngimage01.png

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

image05.png

...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.

image06.pngimage06.png

image06.png

image06.pngimage06.png

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.

image07.pngimage07.png

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.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Number Stairs, Grids and Sequences essays

  1. "Multiply the figures in opposite corners of the square and find the difference between ...

    Firstly a 2 by 2 square I will find the difference. I will randomly place the 2 by 2 square on the grid then find the results. 2 3 13 14 3 x 13 = 39 2 x 14 = 28 Difference is 11 The algebra to back up the

  2. Investigation of diagonal difference.

    the bottom left corner is twice the length of the grid, from noticing this I can now implement G into the equation, only if the number added to n in the bottom left corner is always the size of the grid that the 3 x 3 cutout originated from.

  1. Investigate Borders - a fencing problem.

    square 1x3): Formula = Simplification = As before, I have already done the working out for this Formula and it has been proved. Formula to find the number of squares needed for each border (for square 2x3): Formula = Simplification = Again the working out's been done.

  2. Investigating when pairs of diagonal corners are multiplied and subtracted from each other.

    If you take the previous box size then square it, then multiply by 10 you get the difference on the diagonals of a box size on a 10 x 10 grid. I predict that for a 5 x 5 box on a 10 x 10 grid the difference of the multiplied diagonals will be 160.

  1. I am doing an investigation to look at borders made up after a square ...

    4 4 4 4 5 5 5 5 5 5 Border Number=B Number of numbered squares=N 1 12 2 16 3 20 4 24 5 28 Using my table of results I can work out a rule finding the term-to-term rule.

  2. "Multiply the figures in opposite corners of the square and find the difference between ...

    The number generated will be the number in the top left hand corner of the square. 89 90 99 100 90 x 99 = 8910 89 x 100 = 8900 The difference yet again is 10; this is no mere coincidence and therefore is a constant pattern within all the 2 by 2 squares that I have tested so far.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work