In this investigation I will be trying to find an equation, which will allow me to work out the sum of numbers inside a 'T' shape on a number grid.

Tn = Total number contained within the T.

I will now see if this formula is applicable for any T of this size on a width 10 grid.

Check using formula:

T2 = 1+2+3+12+22 = 40                Tn = (5x2)+30

                                        = 10+30

                                        = 40

This formula works for any T of height 2 and with a top bar of width 3.

Now I will try and find a formula for a T of height 2 and with a top bar of width three but in any size grid:

T3 = 2+3+4+8+13

T3 = 30

Again I will express these numbers in terms of N.

Tn = (N-1)+N+(N+1)+(N+5)+(N+10)

Tn = 5N+15

I will now compare the two equations that I have for different grid sizes:

Size 5                = 5N+15

Size 10                = 5N+30

From external research I have also worked out that a grid with width 15 has the equation 5N+45.

From these results it is easy to spot a pattern, that pattern is that the number that must be added on to the end of 5N is always 3 times the width of the grid. To add this into my equation I will need a new letter, I will use W, this meaning the width of the grid.

The equation would now be:

Tn = 5N+3W

To be sure I will check using the grid width of 15:

Tn = 8+9+10+24+39                        Using equation: (5x9)+(3x15)

= 90                                                           45+45

                                                           = 90

This new formula that includes two variables works.

For my third variable I am going to try and modify my formula to include a different width of the top bar in the T.

Before this I had also tried rotating the T however this failed as it involved making up a completely new equation which could not be entered into the equation containing width of the grid and the T of height 2 and with a top bar of width 3.

So the first thing that I must do is to draw a new T with a wider top bar, I am going to use a T with a top bar of width 5.

This time:

T3 = 1+2+3+4+5+8+13

T3 = 35

So:

Tn = (N-2)+(N-1)+N+(N+1)+(N+2)+(N+5)+(N+10)

Tn = 7N+15

Note that 15 is 3W

Try with grid 10:

T6 = 72

Use formula:

T6 = (7x6)+(3x10)

T6 = 72

The formula Tn = 7N+3W would appear to work for any size grid.

This formula however now needs to be inserted into the original formula.

Again I will need a new letter, from my original formula I can see using B for the number of boxes along the top row the formulae for:

3B = 5N+3W

5B = 7N+3W

Again from further research I found out that 7B = 9N+3W

So again a pattern has emerged quite quickly and it is clear that when the number of boxes goes up by an extra box on each side the number in N rises by 10.

So I can write the formula as N(B+2)+3W = Tn

I must now check this formula on my original grid from the first page:

T5 = 55                                Using formula: 5(3+2)+(3x10)

                                                        (5x5)+30=55

                                                        25+30 = 55.

This formula appears to work.

So my new formula is N(B+2)+3W=Tn

For my fourth variable I will try and insert a letter into my equation, which relates to the height of the T.

In this coursework the height of the T is the number of boxes in the vertical section of the T excluding the box, which is also a part of the horizontal section.

Below is an example of what I class as the height of the T, so all the t’s that I have used have been of height 2.

1        2        3

        5        

        8

First I will try this in a grid of width 5.

So I will now express the numbers in terms of N:

Then Tn = (N-1)+N+(N+1)+(N+5)+(N+10)+(N+15)+(N+20)

         Tn = 7N+50

I will now check this formula using grid width 10:

T4 = 128        Use formula: (7x4)+50

                        =78

This formula does not work and upon further consideration I am going to try a completely new method, this is by making a table and looking for a pattern.

                

I believe that the solutions using the formula are all the same because my formula does not include the height of the T. I will try replacing the 2 in the (b+2) section of the formula with the height.

I will now try this for the first 3 heights.

3(3+1)+30=42

The answer 42 is too large by 20; the answer should come out as 22. To make this work the 3W would have to change to 1W. I will now try this with the second height.

3(3+2)+10=25

The answer 25 is too small by 20, the answer should be 45. To get this to work the 1W would have to change back to 3W. I will try this again with the next height:

3(3+3)+30= 48

The answer 48 is again too small, this time too 30, the answer should be 78. For this too work the W number would have to be increased to 6.

However I have now spotted a pattern, which is needed to make each W number work. The numbers are 1,3,6… These are the triangular numbers. This makes the height variable easy to add into my final equation:

N(B+H)+∴W

Unfortunately I cannot leave a shape in my equation, however after doing some research the equation for a triangular number is:

T(T+1)

     2

So if I insert this into my final equation it will look like this:

Tn = N(B+H)+(H+1)W

        2

Here is a summary of all the letters:

N= T number

Tn = Total of all numbers inside the T

B = Total number of boxes along the top row of the T

W = The width of the grid

H = Height of the T

I will now check my final formula using my original grid.

T3 = 45        Use Formula:  3(3+2)+(2+2)x10

                                         2

                        15+  30 = 45

THE OVERALL FORMULA WORKS!