Maths Coursework- Borders

n = 1  ⇒    1 = a   +   b + 5      1

n = 2  ⇒    1 = 4a + 2b + 5      2   

1   × 2    2a + 2b = - 4

2   × 1    4a + 2b = - 4

1    -  2        - 2a = - 4

                       a = - 4

                               2

                       a = 2

Substitute a = 2 in 1    2 + b = - 4    

                                             b = - 6

 nth term = 2n2 – 6n + 5

White Squares;

1. ,  4  ,   8  ,  12  ,  16  ,  20  ,  24  

4   ,  4  ,   4  ,   4  ,   4  ,    4

nth term = 4n – 4    

                           because the shape adds on 4 white squares each new pattern. It is easy to see the rule, as it is constant and linear. You get 4n as the sequence starts on 0 white squares and not 4 white squares so – 4 every time.

All Squares;

    zero term  →  1     1  ,  5  ,  13  ,  25  ,  41  ,  61  ,  85

1.  4  ,  8  ,  12  ,  16  ,  20  ,  24

4. 4  ,  4  ,  4  ,  4  ,  4  ,  4

nth term = an2 + bn + c

                                         n = 1 ⇒   1 =   a +   b + 1      1

                                         n = 2 ⇒   5 = 4a + 2b + 1      2

                                         

                                         1   ×  2    2a + 2b = 0

                                         2   ×  1    4a + 2b = - 4

                                         

                                         1   -   2    -2a = - 4

                                                           a = 2

                                       

                                        Substitute in a = 2 in 1     2 + b = 0

                                                                                         b = - 2

                                         nth term = 2n2 – 2n + 1

Testing the formulae

Now I must test these rules to see if they are correct. First I will test them on a pattern I have already done and then on a pattern I don’t know the answer to.

Using pattern 3

Total no. of squares;

                                       2n2 – 2n + 1

                                   2 × 32 – (2 × 3) + 1 = 18 – 6 + 1

                                                                  = 13

White squares;

                                   4n – 4

                                   (4 × 3) – 4 = 12 – 4

                                                    = 8

Black squares;

                                              2n2 – 6n + 5

                                  2 × 32 – (6 × 3) + 5 = 18 – 18 + 5

                                                                 = 5

             

This shows that the formulae I have used are correct as the equations give the correct results for cross-shape pattern 3.

Now I will test the formulae on pattern 9 - a cross-shape I don’t know the answer to.

 Total no. of squares;

                                       2n2 – 2n + 1

                                   2 × 92 – (2 × 9) + 1 = 162 – 18 + 1

                                                                  = 145

White squares;

                                   4n – 4

                                   (4 × 9) – 4 = 36 – 4

                                                    = 32

Black squares;

                                              2n2 – 6n + 5

                                  2 × 92 – (6 × 9) + 5 = 162 – 54 + 5

                                                                 = 113

                                                           

I must draw pattern 9 to check that the formulae have worked and I have reached the correct results.

             

Justification of the 2D cross-shapes and formulae

I must look at the cross-shape in a different way, by tilting the shape and using lines I can clearly see and work out how many white squares there are.

Using pattern 3 – to find white squares

                            there are 4 sides with 3 squares on each side so..

                   

White squares;            3 × 4 = 12 – 4 ( - 4 as corners count twice )

                                    4 × 3 (n) = 12

                                    4 × 3 (n) – 4 = 8

same as 4n – 4 and 4 × ( n –1 )

Using pattern 4 –  to find white squares

Using the rule 4n – 4     so…=  4 × 4 – 4   ( - 4 as corners are counted twice )

                                              = 16 – 4 = 12

also using the rule 4 × ( n – 1 )  so… = 4 × ( 4 – 1 )

                                                           = 4 × 3

                                                           = 12

This shows how and why the nth term for the justification works.

Using pattern 3 – to find total number of squares

We need to use a quadratic rule because when the sequence is written out and the difference is worked out, the second differences are equal, which gives you n2 used in a quadratic equation.

White = 32 

Black = 22                           total no. of squares = 32 + 22 = 13

So… for size n, total = n2 + ( n - 1 ) 2

Which is the same as my previous rule for total number of squares as it gives the same result.

This is another way for finding an nth term for the total number of square using cross-shape 4.

Part 2 -  Extending Investigation into 3 dimensions.

I am trying to find out how many cubes it takes to make any 3 dimensional shape built up in this way.

Each 3D cross-shape is made by adding another layer of cubes onto the existing shape.

The first 3D cross-shape will be made out of 1 cube.

To start my investigation I must attempt to draw the first four 3D cross-shapes. I can then record the total number of cubes used to make each 3D cross-shape, from this I should be able to see a pattern in my sequences this should then help me in working out a formula to find any 3D cross-shape built up in this way. I must then try and test the formula using a 3D cross-shape I know and one I don’t know.  

I will then justify my formula by explaining how it works and other ways of reaching the same answer.

Drawing the 3D cross-shapes

Pattern 1                          Pattern 2                             Pattern 3     

Pattern 4

From my drawings of the cross-shapes in 3 dimensions, it is possible for the total number of squares for each pattern to be recorded into a table;

Working out the formula

From the table we get a pattern sequence that looks likes this;

              zero term         - 1  ,  1  ,  7  ,  25  ,  63  ,  129  ,  231

                                           2  ,  6  ,  18  ,  38  ,  66  ,  102

                                               4   ,  12  ,  20  ,  28  ,  36

                                                     8  ,  8  ,  8  ,  8

nth term =  an3 + bn2 + cn + d                    d = zero term

n = 1                 a + b + c –1 = 1

n = 2                 8a + 4b + 2c –1 = 7

n = 3                 27a + 9b + 3c –1 = 25

a + b + c = 2                1

8a + 9b + 2c = 8          2

27a + 9b + 3c = 26      3

 

2  - ( 2 × 1 )   =    _ 8a + 4b +2c = 8

                              2a + 2b +2c = 4

                     =       6a + 2b        = 4             4

3  - ( 3 × 1 )   =    _ 27a +9b + 3c = 26

                               3a + 3b + 3c =  6

                      =        24a + 6b      = 20         5

Multiply equation 4 by 3 so that 6b can be cancelled out of the equation and we can find out what a is equal to,

4 × 3   =  18a + 6b  =  12

Then to find a we need to subtract equation 5 from equation 4

4 – 5   =   _18a + 6b  = 12

                  24a + 6b  =  20

                         - 6a   =  -8

                             a   =  8  =  4                                      

                                      6      3

Substitute in   a = 4 in 4

                             3                    6a +  2b  =   4

8   +  2b  =   4

          2b  = - 4

           b  = - 2

 Substitute in  b = - 2  in 1

                                                    4 – 2 + c = 2

                                                    3        

                    c = 4  –  4

                                   3  

                c = 2  2

                                                                                                               3

                                                                c = 8

                 3

I can now place these values in an equation for finding the nth term of any 3 dimensional cross-shape built up in this way.

nth term = 4 n3   -  2n2  +  8 n  - 1

3. 3

Testing the formula

I will test the formula on two cross-shape patterns in order to determine whether the formula achieved is correct.

Using 3D cross-shape pattern 2

nth term = 4 n3   -  2n2  +  8 n  - 1

                 3                       3

              = (  4 × 23 )  - ( 2 × 22 ) + (  8 × 2  ) - 1

3.                                 3

              =  7

The formula tells us that 3D cross-shape pattern 2 would have 7 squares, this is the same answer as in the table, showing us that the formula being used is correct.

Next I will try…

Using 3D cross-shape pattern 6

nth term = 4 n3   -  2n2  +  8 n  - 1

                 3                     3

             = (  4 × 63 )  - ( 2 × 62 ) + (  8 × 6  ) - 1

3. 3

             =  231

This answer also corresponds correctly with the answer in the table which indicates that the formula is successful in finding any 3D cross-shape pattern built up in this way.

Justification of the 3D cross-shapes

I will now explain why my formula works and also different ways of coming to the same answer using a different method.

3D pattern 2                                                    3D pattern 3

 

If you look at the 3D shapes from a side view they’d look like this.

Each layer corresponds with the 2D cross-shape patterns,

for example layer 1 in the 3D cross-shape corresponds with the 2D cross-shape pattern 1 as they both have the same total number of squares, same goes for layer 2 in the 3D cross-shape as this corresponds with 2D cross-shape pattern 2 also having the same number of squares.

This is how you can work out the total number of squares more easily as you can use the answers to the 2D shapes to help you as the 3D shapes are hard to draw.

Another formula possible to find the total number of squares for any 3D shape built up in this way is;

2 ( layer 1 + layer 2 + layer 3…) + layer n

 

or

• ( 2n2 – 2n +1 ) + 2 ( 2 ∑ r2 – 2 ∑ r + n )

By using the appropriate algebra this equation should lead to the appropriate answer.

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