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

1st Diff:

                    4                   8                    12                   16                   20

2nd Diff:

                              4                     4                     4                     4

To find the 1st and  2nd difference, I used a table and then the diagram above to check if I was right about my suggested answers. First of all, for the 1st difference, I took the number of squares for shape number 2 and subtracted that with the number of squares I got for the first shape, which then gave me the 1st difference. I continued like this until I had all my 1st differences.

Afterwards, I took the result from the 1st difference of shape number 2 and subtracted that with the 1st difference of shape number 1, I repeated this until the end, and the 2nd 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

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 

2. 612= 72

From there, to get the 1st differences, I took the remaining number of the second shape and subtracted that to the remaining number of the first shape. I continued until I had all the same results which was -2.

The formula I was to investigate was:

an    +    bn    +    c

This would become:

2n   +    -2n    +   1

To testify this formula I will draw a 7th cross-shape  to test that:

n   =    7

This would therefore become:

2 ( 7  )  -   2 ( 7 + 1 )  =   85

At the end I'll make sure that I drew the right amount of squares for my shape by counting appropriately.

Table 3

As I have had 3 sets of differences( this is called a Cubic Function which means anything to the power of 3 ) before the differences remained constant, I now know that my formula will n3 in it.

The 3rd difference is therefore 8 which means:

8. a= 8

                       6

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

6. 3           

I will now use 4       n2 to make a sequence.

                              3

The second part of the equation will now be:

-6       n2      =      -2n2

      3

I will use this to make another sequence.

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.