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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

From evidence in the table, the formula for the number for the number of white squares can be found as it is just an arithmetic progression, the number increasing by 4 each time. This gives the formula to find the number of white surrounding squares as:

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

12

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    1. 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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