# Borders - Investigation into how many squares in total, grey and white inclusive, would be needed to make any cross-shape of this kind

Extracts from this document...

Introduction

Borders

This investigation involves diamonds of grey squares, surrounded by a number of white ones, for example:

And the initial investigation is into how many squares in total, grey and white inclusive, would be needed to make any cross-shape of this kind.

To start with for this investigation, I drew out the first four possible build up of these cross-shapes and recorded the results in a table:

Width: 1 Grey SquareWidth: 5 Grey Squares

## Width: 3 Grey SquaresWidth: 7 Grey Squares

Width / Grey | Total / Shaded | Total / White | Total / All |

1 | 1 | 4 | 5 |

3 | 5 | 8 | 13 |

5 | 13 | 12 | 25 |

7 | 25 | 16 | 41 |

This table shows there is a relationship between each of the set of numbers produced.

As the width increases by two each time, letting the width = n means that:

N1 = 1, N2 = 3, N3 = 5 and N4 = 7

This relation between which term applies to a width can be shown as:

Nno.= Width + 0.5

2

From hereon, each suffix will have been derived from the width of shaded squares, in the way mentioned above.

Next, looking at the increases between the totals of both shaded and white

Middle

Un= 4n

Following on from this, the total number of squares, with the total of white surrounding squares removed, gives the total of shaded squares for each width.

Un = 2n² + 2n + 1 – 4n

= 2n² - 2n + 1

And this result is the same as the previous total number of squares, the size and shape of the cross increasing regularly:

5/2 + 0.5 = 3 3/2 + 0.5 = 2

18 – 6 +1 = 13 8 + 4 + 1 = 13

From this I can find out if my formula for the total number of squares is correct by looking at two different ways of getting the next term in the series:

2(n + 1)²+ 2(n + 1) + 1 must = 2n² + 2n + 1 + 4(n + 1)

LHS = 2(n² + 2n + 1) + 2n + 2 + 1

=2n² + 4n + 2 + 2n + 2 + 1

= 2n² + 6n + 5

RHS = 2n² + 2n + 1 + 4n + 4

=2n² + 6n + 5

And this proves that the formula is correct and works for all values of n.

Conclusion

U7 – U9 = 42a + 6b = 68 = U11

U11 – U12 = 6a = 8

- a = 4/3

(4/3 x 12) + 2b = 20

- 16 + 2b = 20
- 2b = 4
- b = 2

(4/3 x 7) + (2 x 3) + c = 18

- 46/3 + c = 18
- c = 8/3

4/3 + 2 + 8/3 + d = 7

- 6 + d = 7
- d = 1

This gives the formula:

Un = 4n³+ 6n²+ 8n + 3

3

To prove this formula, I need to find the formula for finding the number of white squares so I can find the next term in the series two ways, and therefore prove the formula works:

6 18 38 66

12 20 28

- 8

This shows there is a squared relation between these totals, and the simultaneous equations are used again.

Un = an² + bn + c

U1 = a + b + c = 6

U2 = 4a + 2b + c = 18

U3 = 9a + 3b + c = 38

U2 – U1 = 3a + b = 12 = U4 U4 x2 = 6a + 2b = 24 = U6

U+3– U1 = 8a + 2b = 32 = U5

U5 – U6 = 2a = 8

- a = 4

(3 x 4) + b = 12

- b = 0

4 + 0 + c = 6

- c = 2

Giving the formula:

U7 = 4n² + 6

From this a proof of the formula for Un can be found:

4(n+1)³+ 6(n+1)²+ 8(n+1) + 3 must = 4n³+6n²+ 8n + 3 + 12(n+1)²+ 18

3 3 3

## LHS = 4(n+1)(n+1)²+ 6(n²+2n+1) + 8n +11

3

= 4(n+1)(n²+2n+1) + 6n²+12n+6 + 8n + 11

3

= 4(n3+2n2+n+n2+2n+1) + 6n2 + 12n + 6 +8n +11

3

= 4n3 + 8n2 + 4n + 4n2 + 8n + 4 + 6n2 + 12n + 6 + 8n +11

3

= 4n3 + 18n2 + 32n + 21

3

RHS = 4n³+6n²+ 8n + 3 + 12(n+1)²+ 18

3

= 4n3 + 6n2 + 8n + 3 + 12 (n2+2n+1) + 18

3

= 4n3 + 6n2 + 8n + 3 + 12n2 + 24n + 18

3

= 4n3 + 18n2 + 32n + 21

3

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