From these tables, it is possible to see a useful pattern:
- The Sum of the Tail equals the Middle Number plus the Grid Width.
5) Generalisation
It can be assumed that for all possible locations of the 3x1 “T” on the width g grid, these patterns will be true. Therefore, the following logic can be used to create a formula where:
n = the Middle Number
g = the Grid Width
-
Using Pattern 1 above, we can say that the Sum of the Tail = n + g
-
Using the patterns from Section 1, we can still say that the Sum of the Wing = 3n
-
Because Total Sum = Sum of Wing + Sum of Tail, then Total Sum = 3n + n + g
Total Sum = 4n + g
This means I predict that with a 3x1 “T” on a width g grid, the sum will always be 4n + g.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where n = 20 and g = 15
2) Where n = 63 and g = 17
7) Justification
The formula can be proven to work with algebra. The basic algebraic labelling of the 3x1 “T” on a width g grid is:
n is the Middle Number;
(n + 1) is the top-right number in the “T”, because it is always “1 more” than n;
(n - 1) is the top-left number in the “T”, because it is always “1 less” than n;
(n + g) is the tail, because it is always “g more” than n.
In the “4n + g” formula, the “4” represents the number of “n”s which are present within the “T”. Every square in the “T” is referred to in terms of n and so with 4 squares added together, there are 4n.
The “g” comes from the “Sum of the Tail”, which is g more than the Middle Number.
The “Sum of the Wing” is 3n because there are 3 “n”s present in the 3 squares of the wing. The “-1” and “+1” cancel out each other when added to together.
8) Conclusion
After this justification, it can now be said that for every possible 3x1 “T” on a Width g Grid, the Total Sum of all of the squares contained within it is 4n + g.
9) Extension
Having done this, I saw that my formula would only work for a 3x1 “T”. To improve the usefulness of my formula, I wondered what would happen to the Total Sum if I varied the Width of the Wing of the “T” i.e. a wx1 “T”.
Section 3: w x 1 “T” on Width 20 Grid
1) Introduction
Throughout this section, the variable w will be used to represent the width of the “wing” of the “T”. The variable n will continue to be used for the Middle Number in the “T” i.e. the location of the “T” upon the grid.
2) Method
Varying values of w will be tested to give different widths of wings for the “T”. The widths will be 5, 7 and 9. Only odd numbers can be used for the wing width, because the “T” always has an equal number of boxes either side of the Middle Number. With these "T”s, in 5 different locations on the width 20 grid, the Total Sum will be calculated.
Width 20 must be used for this section because when varying the widths of “T”, the grid must be wide enough to give a good range of values for data collection, and later testing.
3) Data Collection
Fig 3.1
Fig 3.1 is the width 20 grid, with 3x1, 5x1, 7x1 and 9x1 example “T”s.
a) Here are the results of the 5 calculations for a 5x1 “T” on Width 20 Grid:
b) Here are the results of the 5 calculations for a 7x1 “T” on Width 20 Grid:
c) Here are the results of the 5 calculations for a 9x1 “T” on Width 20 Grid:
4) Data Analysis
From the tables (a)-(c), it is possible to see that when only the wing width is varied, only the Sum of the Wing changes. In fact, if we take one constant Middle Number, 25, from each of the above tables, we get the following:
From these tables, it is possible to see a useful pattern:
- The Sum of the Wing equals the Middle Number multiplied by the Wing Width.
- The Sum of the Tail equals the Middle Number plus 20.
5) Generalisation
It can be assumed that for all possible locations of the wx1 “T” on the width 20 grid, these patterns will be true. Therefore, the following logic can be used to create a formula where:
n = the Middle Number
w = the Wing Width
-
Using Pattern 1 above, we can say that the Sum of the Wing = wn
-
Using Pattern 2 above and knowledge from Section 1, we can say that the Sum of the Tail = n + 20
-
Because Total Sum = Sum of Wing + Sum of Tail, then Total Sum = wn + n + 20
If this is factorised, we get the following:
Total Sum = (w + 1)n + 20
This means I predict that with a wx1 “T” on a width 20 grid, the sum will always be (w + 1)n + 20.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where n = 92 and w = 13
2) Where n = 128 and w = 15
7) Justification
The formula can be proven to work with algebra in the form of an arithmetic progression. The basic algebraic labelling of the wx1 “T” on a width 20 grid is:
For the purposes of simplicity, p = which represents the last number which is added or subtracted in the series
either side of the Middle Number, as illustrated above.
n is the Middle Number;
(n + p) is the top-right number in the “T”, because it is always “p more” than n;
(n - p) is the top-left number in the “T”, because it is always “p less” than n;
(n + 20) is the tail, because it is always “20 more” than n.
In the formula, the “(w + 1)” represents the number of “n”s which are present within the “T”. Every square in the “T” is referred to in terms of n and so with w number of squares in the Wing and “1” in the Tail, the number of “n”s is (w + 1).
The “20” comes from the “Sum of the Tail”, which is 20 more than the Middle Number.
The “Sum of the Wing” is wn because w number of “n”s are present in the w number of squares in the wing. Because equal but opposite values are present on either side of the Middle Number, all the numbers cancel each other out to leave just “n”s.
8) Conclusion
After this justification, it can now be said that for every possible wx1 “T” on a Width 20 Grid, the Total Sum of all of the squares contained within it is (w + 1)n + 20.
9) Extension
Having done this, I saw that my formula would only work for varying the Width of the Wing of the “T”. To improve the usefulness of my formula, I wondered what would happen to the Total Sum if I varied the Length of the Tail on the “T” i.e. a 3 x l “T”.
Section 4: 3 x l “T” on Width 10 Grid
1) Introduction
Throughout this section, the variable l will be used to represent the length of the “tail” of the “T”. The variable n will continue to be used for the Middle Number in the “T” i.e. the location of the “T” upon the grid.
2) Method
Varying values of l will be tested to give different lengths of tails for the “T”. The lengths will range from 1 to 5. With these “T”s, in 5 consecutive locations on the width 10 grid, the Total Sum will be calculated.
Width 10 will be used for this section because it is a number with which patterns are easier to spot.
3) Data Collection
Fig 4.1
Fig 4.1 is the Width 10 grid, with 3x1, 3x2, 3x3, 3x4 and 3x5 example “T”s.
a) Here are the results of the 5 calculations for a 3x1 “T” on Width 10 Grid:
b) Here are the results of the 5 calculations for a 3x2 “T” on Width 10 Grid:
c) Here are the results of the 5 calculations for a 3x3 “T” on Width 10 Grid:
d) Here are the results of the 5 calculations for a 3x4 “T” on Width 10 Grid:
e) Here are the results of the 5 calculations for a 3x5 “T” on Width 10 Grid:
4) Data Analysis
From the tables (a)-(e), it is possible to see that when only the tail length is varied, only the Sum of the Tail changes. In fact, if we take one constant Middle Number, 15, from each of the above tables, we get the following:
The columns that are of interest in this table are the Tail Length and the Sum of the Tail. In order to spot a relationship between the two, the Sum of the Tail must be expanded into all of its original terms. See overleaf.
This table shows all the Sums of the Tails, with the Middle Number 15 and the Grid Width 10:
Fig 4.2
From Fig 4.2, it is possible to see a very useful pattern:
- The Sum of the Tail divided by the Tail Length equals the mean of the numbers in the Tail Boxes:
Alternatively, this means that the Sum of the Tail equals the Mean of the Tail Boxes multiplied by the Tail Length:
-
Also, the last number (at the bottom of each column of the Tail Boxes row in Fig 4.2) seems to be the Middle Number (15), plus 10, 20, 30 etc. which are all multiples of 10. In fact, the last number seems to be the Middle Number plus the Tail Length x 10.
5) Generalisation
It can be assumed that for all possible locations of the 3xl “T” on the width 10 grid, these patterns will be true. Therefore, the following logic can be used to create a formula where:
n = the Middle Number
l = the Tail Length
-
Using Pattern 1 above, we can say that the Sum of the Tail = Mean[Tail Boxes] x l.
However, the “Mean of the Tail Boxes” must be expressed in terms of l and n in order to create a formula.
The mean of a set of numbers which increment by a constant amount each time can be found by summing the first and last numbers in the sequence, and then dividing by 2. An example of this can be seen if we take the “5” column from Fig 4.2:
Using Pattern 2 above, we can say that the first term in the sequence of our numbers is the Middle Number plus the Tail Length x 10, or n + 10l. Since l is 1 in the first term of the sequence, the first term is therefore n + 10.
The last number in the sequence can be found using Pattern 2 again, and hence it is n + 10l. This is because the length of the tail denotes where the last number in the sequence can be found.
And so now, if we “sum the first and last numbers in the sequence and divide by 2”, we should get the mean of the set of numbers in the tail. Algebraically, therefore:
Finally, using Pattern 1 from above, we say that Sum of Tail = Mean[Tail Boxes] x Tail Length, and so algebraically again, we get this:
To complete this section, the Sum of the Wing must be added to the Sum of the Tail to get the Total Sum. In Section 1, it was found that with a Width 3 wing, the Sum of the Wing was 3n. This will therefore be used again to create the formula:
Total Sum = 3n + ½ l {2n + 10(l + 1)}
This means I predict that with a 3xl “T” on a width 10 grid, the sum will always be 3n + ½ l {2n + 10(l + 1)}.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where n = 32 and l = 7
2) Where n = 29 and l = 8
7) Justification
The formula can be proven to work with algebra in the form of an arithmetic progression. The basic algebraic labelling of the 3xl “T” on a width 10 grid is:
n is the Middle Number;
(n + 1) is the top-right number in the “T”, because it is always “1 more” than n;
(n - 1) is the top-left number in the “T”, because it is always “1 less” than n;
(n + 10) is the First Term in the Tail, because it is always “10 more” than n.
(n + 10l) is the Last Term in the Tail because there are always “l” number of “10”s in it.
Now, using the already justified “3n” that was found in Section 1, the Total Sum formula can be created:
The first step was to add the First Term to the Last Term, the 2nd Term to the 2nd to Last Term etc. in order to show that these pairs of numbers sum to make one constant value.
Having found the constant value that is particular to this series (i.e. the grid width, the middle number and the tail length), when divided by 2, it gives the mean of the series of numbers. Because a mean is the sum of a set of numbers, divided by the number of terms in that set, we can find the sum of the set by multiplying the mean by the number of terms:
The “½{2n + 10(l + 1)}” part of the formula gives us the mean, and so this is multiplied by l because l is the length of the tail, and therefore the number of boxes (or terms) which are present in the tail. This therefore produces the sum.
The sum of the wing has already been justified in Section 1, and the explanation is the same since no changes have been made which affect it.
8) Conclusion
After this justification, it can now be said that for every possible 3xl “T” on a Width 10 Grid, the Total Sum of all of the squares contained within it is 3n + ½ l {2n + 10(l + 1)}.
9) Extension
Having done this, I saw that my formula would only work for varying the Length of the Tail of the “T” on its own. To improve the usefulness of my formula, I wondered what would happen to the Total Sum if I varied the Length of the Tail on the “T and the Width of the Wing and the Grid Width simultaneously i.e. a w x l “T” on a Width g Grid.
Section 5: w x l “T” on Width g Grid
1) Introduction
Throughout this section, the variable w will be used to represent the width of the “wing” of the “T” and the variable l will be used to represent the length of the “tail” of the “T”. The variable g will be used to represent the grid width and the variable n will continue to be used for the Middle Number in the “T” i.e. the location of the “T” upon the grid.
2) Method
Varying values of l, w and g will be tested to give different lengths of tails and widths of wings for the “T”, and change the grid width. With these T”s, in 5 consecutive locations on varying widths of grid, the Total Sum will be calculated.
Because of the large amount of combinations of variables, the method for this will simply to be to use several combinations. Since all these variables have been tested before, it should be easy to spot the earlier patterns using previous knowledge that should remain true.
3) Data Collection
Fig 5.1
Fig 5.1 is an example Width 10 grid, with example “T”s ranging from 3x1 to 9x5. For examples of different grid widths, see Figs 2.1-2.5.
a) Here are the results of the 5 calculations for a 5x1 “T” on Width 10 Grid:
b) Here are the results of the 5 calculations for a 5x3 “T” on Width 10 Grid:
c) Here are the results of the 5 calculations for a 7x4 “T” on Width 11 Grid:
d) Here are the results of the 5 calculations for a 9x6 “T” on Width 12 Grid:
e) Here are the results of the 5 calculations for a 11x7 “T” on Width 13 Grid:
4) Data Analysis
The aim of this analyis is to find the patterns that were discovered in the other sections of the investigation.
A) The Grid Width Pattern
The difference between Section 1 and Section 2 was that the formula changed from “4n + 10” to “4n + g”. This assumes that in place of the “10”, any grid width could be used and the “10” had to be replaced with a “g”. This assumption will be used in the analysis.
B) The Wing Width Pattern
In Section 3, the Wing Width was tested to find a pattern. The pattern found was that the Sum of the Wing = Middle Number multiplied by Wing Width. We can see that this is true by taking a few examples from the tables above:
As a result of this, the pattern that Sum of Wing = wn can be deemed to be true.
C) The Tail Length Pattern
In Section 4, the Tail Length was tested to find a pattern. The pattern found was that:
Sum of Tail = ½ l {2n + 10(l + 1)}
From (A), in this section, the formula has been hypothesised to work when the “10” is replaced with the Grid Width. In order to test this hypothesis, a new formula has been created for the Sum of Tail that includes the variable g:
Sum of Tail = ½ l {2n + g(l + 1)}
Although this was not found by the methods in other sections, for the purposes of spotting patterns, the use of knowledge from other sections is not detrimental to the investigation. Here is a table that shows this formula to apparently work, fulfilling the pattern from the previous section:
As a result of this, the pattern that Sum of Tail = ½ l {2n + g(l + 1)} can be deemed to be true, but needs to be tested and justified.
5) Generalisation
It can be assumed that for all possible locations of the wxl “T” on the width g grid, these patterns will be true. Therefore, the following logic can be used to create a formula where:
n = the Middle Number
w = the Wing Width
l = the Tail Length
g = the Grid Width
By using the formulas created from previous knowledge in the Data Analysis part, the following formula can created for the Total Sum:
Total Sum = wn + ½ l {2n + g(l + 1)}
This is a combination of the formulas for the Sum of the Wing and the Sum of the Tail on any grid width that were created in the Data Analysis.
This means I predict that with a wxl “T” on a width g grid, the sum will always be wn + ½ l {2n + g(l + 1)}.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where n = 177, w = 23, l = 14, g = 40
2) Where n = 115, w = 27, l = 17, g = 50
7) Justification
The formula can be proven to work with algebra in the form of an arithmetic progression. The basic algebraic labelling of the wxl “T” on a width g grid is:
For the purposes of simplicity, p = which represents the last number which is added or subtracted in the series
either side of the Middle Number, as illustrated above.
n is the Middle Number;
(n + p) is the top-right number in the “T”, because it is always “p more” than n;
(n - p) is the top-left number in the “T”, because it is always “p less” than n;
(n + g) is the First Term in the Tail, because it is always “g more” than n.
(n + gl) is the Last Term in the Tail because there are always “l” number of “g”s in it.
A) Justification of Sum of Wing
This is really the same justification that was in Section 3, but is repeated for clarification here:
B) Justification of Sum of Tail
This is the same justification as in Section 4, but the “10”s have been replaced with “g”s.
Now, using the already justified “wn”, the Total Sum formula can be created:
8) Conclusion
After this justification, it can now be said that for every possible wxl “T” on a Width g Grid, the Total Sum of all of the squares contained within it is wn + ½ l {2n + g(l + 1)}.
Investigation Conclusion and Evaluation
From the simple study of a 3x1 “T” on a width 8 grid, I have been able to progress all the way to the formula to find the sum of the “T” on any width grid. Patterns were easy to spot early on, however when advancing into more complex areas of the project, like the last section where three different variables were altered, spotting patterns became more difficult and knowledge from the prior sections was required to find the formula.
Overall, the best way of presenting the results was in tables because there was an obvious pattern of values in the “Difference” column, because they were all in one line. A graph or diagram would not have been as suitable because the patterns would not have been as apparent, except in a linear progression but this is easily seen in a pattern as the numbers incrementing by a constant amount each time.
There are limitations on the formula though. The width of the wing must be never create a situation where any boxes are outside the grid. This relies on the middle number being chosen carefully to ensure, even if the wing width appears it will fit on the grid (like a Wing Width of 8 on a Width 50 Grid), that it is not so close to the edge that the “T” is not wholly on the grid.
The final justified formula is that the Total Sum =
wn + ½ l {2n + g(l + 1)}