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This coursework will be to investigate to see how many squares would be needed to make any cross-shape build up in this way.

Extracts from this document...

Introduction

Introduction

The figure below shows a dark cross-shape that has been surrounded by white squares to create a bigger cross- shape:

The bigger cross- shape consists of 25 small squares in total. The next cross- shape is always made by surrounding the previous cross-shape with small squares.

This coursework will be to investigate to see how many squares would be needed to make any cross-shape build up in this way.

Method

First of all I'm going to work out a sequence which consists of a formula. Then I will draw a maximum of six cross-shapes adding on to each sequence, at the end I will produce one extra to testify the formula.

After creating two tables to work out the first

Middle

nd difference turned out to be 4. This is called a quadratic function, this means when it takes two sets of terms to actually find out what the constant difference is.

In the following table I will be intending to find out what the answers are to 2n2.

Table 2

 Shape Number 1 2 3 4 5 6 Number of Squares 1 5 13 25 41 61 2n2 2 8 18 32 50 72 Remaining sequence -1 -3 -5 -7 -9 -11 1st difference -2 -2 -2 -2 -2 -2

No of squares:

1                   5                   13                  25                  41                 61

Remaining Sequence:

-1                 -3                    -5                   -7                   -9

1st Diff:

-2                   -2                    -2                    -2

To find the remaining sequence, I used the formula 2n2  with the number of squares . So basically when I worked out the answer to 2n2, I subtracted the number of squares by the results of my formula, which gave me the answers to the remaining sequence.

2          12    = 2

2          52    =  8

2          132 =18

2          252=  32

2          412= 50

1. 612= 72

Conclusion

The 3rd difference is therefore 8 which means:

1. a= 8

6

The 1st part of the formula becomes 8     nb  =  4    n3

1. 3

I will now use 4       n2 to make a sequence.

3

 n 1 2 3 4 5 6 4    3  n3 4     3 32      3 158         3 256          3 500          3 864          3 Remaining Sequence= 1 -  4            3-1/3 7 - 32             3-11/3 25- 158                3-33/3 63 - 256                3-67/3 1st Diff -10        3 -22        3 -34        3 2nd Diff -12         3 -12         3 -12       3

The second part of the equation will now be:

-6       n2      =      -2n2

3

I will use this to make another sequence.

 n 1 2 3 4 5 6 -2n2 -2 -8 -18 -32 Remaining Sequence -1/3-(-2)5/3 -11/3-(-8)13/3 -33/3-(-18)21/3 -67/3-(-32)29/3 37/3 1st Diff 8/3 8/3 8/3 8/3

I have now found out the last part of the formula which is :

8   n    -     3

3                 3

The formula is :     4/3n3   -   2n2   +   8/3n   -    3/3

= 4n3    -     6n    +    8n  -   3

3

To testify this formula I'm going to draw squares so that the 3D look can be easily understood.

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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