- Then click STAT, select CALC, then select “5:QuadReg”
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Click 2ND, input L1, then comma, input L2
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Then comma again, click VARS, then select Y-VARS, select “1:Function…”
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Then select “1:Y1”
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The answer is:
Let x be the row number and y the numerator
SIMPLIFYING
SIMPLIFYING
Therefore the general statement representing the numerator in each row is:
The denominators in each row were examined, and what was found was that the denominators started to differ from row four and that the left side was symmetrical to the right side. In order to find the denominator a third variable needs to be introduced to represent the position within the row. I looked at one row at the time, below is the working for row five.
Consequently, using Microsoft Excel, I plotted the relationship between the element number and the denominator. The graph below shows this relationship.
To represent this graph in an equation for the denominator, I used a graphic calculator to derive the equation using the quadratic regression function, following these steps.
- Click STAT, then click “1:Edit”
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Put the element in L1 and the denominators for row 5 on L2
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Then click STAT, select CALC, then select “5:QuadReg”
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Click 2ND, input L1, then comma, input L2
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Then comma again, click VARS, then select Y-VARS, select “1:Function…”
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Then click “1:Y1”
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The answer got is:
Let n be the row number, D the denominator, r the element number and y the numerator in the row (changing the x,y,b,c variables)
As we can see from step 14 above, b is negative and therefore when we substitute the variable n in the equation it is also negative. Additionally, the numerator, y, is substituted for the variable c, as it represent the numerator of the row.
The working below proves the equation right for the second denominator element in the fifth row.
The following working proves the equation right for the first denominator element in the fourth row.
The first way I used to find the sixth and seventh rows is the same trend shown at the very start. Using this trend I already found the sixth row and will continue the trend to find the seventh row.
1 1
To make sure the answers were correct, I used the numerator equation that I found earlier.
6th row numerator:
7th row numerator:
I then made sure all the denominators were correct using the denominator formula.
6th row denominators
1st and 5th denominators
2nd and 4th denominators
3rd denominators:
7th row denominators
1st and 6th denominators
2nd and 5th denominators
3rd and 4th denominators
Hence, the pattern in the 6th row is:
1 1
The pattern in the 7th row is:
1 1
Therefore to find the general statement of , where it is the element in the nth row, starting with r = 0, the numerator and denominator equations found earlier are used. The numerator formula is divided by the denominator formula. The numerator formula can also be written as . Now that we know that we can substitute it in the denominator formula for y. Hence, the equation below represents the general statement of .
To prove that the general statement works correctly, the following example is provided, that clearly proves the general statement right.
Now I will test the validity of the general statement by finding the 8th, 9th and 10th rows using the general statement found earlier.
8th row numerator:
9th row numerator:
10th row numerator:
8th row denominators
1st and 7th denominators
2nd and 6th denominators
3rd and 5th denominators
4th denominator
9th row denominators
1st and 8th denominators
2nd and 7th denominators
3rd and 6th denominators
4th and 5th denominators
10th row denominators
1st and 9th denominators
2nd and 8th denominators
3rd and 7th denominators
4th and 6th denominators
5th denominator
Hence, the 8th row reads as:
1 1
The 9th row reads as:
1 1
The 10th row reads as:
1 1
The equation is also true for the 1’s at the beginning and end of the row. When r = 0, you end up with the numerator over the numerator, which equals to 1. The example below proves this statement true for the 1 at the beginning of the 10th row.
Since the general statement’s answer is a fraction, I initially separated the fraction to find first the numerator equation and later the denominator equation. Therefore I looked at the numerator and denominator separately to find patterns. This is what led me to the general statement, which is simply the numerator equation over the denominator equation.
There are several scope and limitations that arise for this general statement.
Firstly, when I first started my investigation I did not take the first row into account. Since the first row is only composed of a 1 and another 1, the equation will not be true to calculate the first row. As well as when using the numerator equation the second element becomes the first element as the 1’s are counted as the 0 element, for the general statement to work.
Secondly, the variable n, must be greater than 0. This because there cannot be negative rows, as if plugged into the general statement the answer would be positive, when it should be negative in order to be correct.
Thirdly, the variable r, must be equal or greater than 0, as there cannot be 0 or negative elements in a pattern. For the equation to work, the variable r, has to be an integer, fractions will not work.
Lastly, the element number, r, must be smaller then the row number n. As we can see from the table below the element is always one less then the row number. Therefore the element, r, cannot be greater then the row number, n, as a row number can only have a limited amount of elements.
In conclusion, through the use of a straightforward pattern, I was able to find the next 2 rows of the given symmetrical pattern. In addition with the use of quadratic regression I was able to derive a general statement for both the numerators and denominators, and prove it for other rows of the pattern. With the help of these two equations, I finally arrived at my general statement. Lastly, thanks to the deeper understanding of the calculation process, I examined in detail the validity, scope and limitations of my general statement.