In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I don’t know the formula so I will have to work that out.
Firstly I will be finding out the formula for the shortest side.
Length - 3 5 7 9 11
1st difference - 2 2 2 2
The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n.
2n
2 x 1 = 2 (wrong)
There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1.
2n+1
2x1+1=3 (correct)
2n+1
2x2=4
4+1=5 (correct)
From looking at my table of results and carrying out some testing of rules, I noticed that ‘an + n = b’. So I took my formula for ‘a’ (2n + 1) multiplied it by ‘n’ to get ‘2n2 + n’. I then added my other ‘n’ to get ‘2n2 + 2n’. This is a parabola as you can see from the equation and also the graph
I will now test it using the first three terms.
2 x 1² + 2 x 1 = 4
2 x 1 + 2 = 4
2 + 2 = 4
4 = 4
My formula works for the first term; so, I will now check it in the next term.
2 x 2² + 2 x 2 = 12
2 x 4 + 4 = 12
8 + 4 = 12
12 = 12
My formula works for the 2nd term. If it works for the 3rd term I can safely say that
2n² + 2n is the correct formula.
2 x 3² + 2 x 3 = 24
2 x 9 + 6 = 24
18 + 6 = 24
24 = 24
My formula also works for the 3rd term. I am now certain that 2n² + 2n is the correct formula for finding the middle side.
Middle side = 2n² + 2n
I now have the much easier task of finding a formula for the longest side.
From the table of results I know that there is only one difference, which is +1 between the middle and longest side. So:
(Middle side) + 1 = longest side
2n² + 2n + 1 = longest side.
I am very certain that this is the correct formula. I will check anyway using the first three terms:
2n² + 2n + 1 = 5
2 x 1² + 2 x 1 + 1 = 5
2 + 2 + 1 = 5
5 = 5
The formula works for the first term.
2n² + 2n + 1 = 25
2 x 3² + 2 x 3 + 1 = 25
18 + 6 + 1 = 25
25 = 25
The formula also works for the second term
2n² + 2n + 1 = 13
2 x 2² + 2 x 2 + 1 = 13
8 + 4 + 1 = 13
13 = 13
The formula works for all three terms. So… Longest side = 2n² + 2n + 1
Now I will check that
2n + 1
2n² + 2n and
2n² + 2n + 1
Form a Pythagorean triple or in other words a² + b² = c²
a² + b² = c²
This equals:
(2n + 1) ² + (2n² + 2n) ² = (2n² + 2n + 1) ²
If you then put these equations into brackets:
(2n + 1)(2n + 1) + (2n² + 2n)(2n² + 2n) = (2n² + 2n + 1)(2n² + 2n + 1)
If I work this equation out by balancing them in each side and I end up with nothing, then 2n + 1, 2n² + 2n and 2n² + 2n + 1 is a Pythagorean triple.
4n² + 4n² = 4n² + 2n² + 2n²
4n = 2n + 2n
8n³ = 4n³ + 4n³
4n = 4n
1 = 1
I now end up with 0 = 0, so 2n + 1, 2n² + 2n and 2n² + 2n + 1 is a Pythagorean triple.
I now have the nth term for each of the three sides of a right-angled triangle. I can now work out, both, the nth term for the perimeter and the nth term for the area.
The perimeter of any triangle is just the length of the 3 sides added together. E.g.
1st term 3 + 4 + 5 = 12
And so on. All I have to do is put all the 3 formulas together.
Perimeter = (shortest side) + (middle side) + (longest side)
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 6n + 2
= Perimeter
If I have done my calculations properly then I should have the right answer. To check this I am going use the 5th and 6th terms.
5th term:
4n² + 6n² + 2 = perimeter
4 x 5² + 6 x 2 = 11 + 60 + 61
100 + 30 + 2 = 132
132 = 132
And it works for the 5th term
And finally the 6th term:
4n² + 6n² + 2 = perimeter
4 x 6² + 6 x 6 + 2 = 13 + 84 + 85
144 + 36 + 2 = 182
182 = 182
It works for all the terms so:
Perimeter = 4n² + 6n + 2
I know that the area of a triangle is found by:
Area = ½ (b x h)
b = base
h = height
Depending on which way the right angled triangle is the shortest or middle side can either be the base or height because it doesn’t really matter which way round they go, as I’ll get the same answer either way.
Area = ½ (shortest side) X (middle side)
= ½ (2n + 1) x (2n² + 2n)
= (2n + 1)(2n² + 2n)
I will check this formula on the first two terms:
(2n + 1)(2n² + 2n) = ½ (b x h)
(2 x 1 + 1)(2 x 1² + 2 x 1) = ½ x 3 x 4
3 x 4 = ½ x 12
12 = 6
6 = 6
2nd term:
(2n + 1)(2n² + 2n) = ½ b h
(2 x 2 + 1)(2 x 2² + 2 x 2) = ½ x 5 x 12
5 x 12 = ½ x 60
60 = 30
30 = 30
It works for both of the terms. This means:
Area = (2n + 1)(2n² + 2n)
To check that these are all Pythagorean triples-
a2 + b2= c2
(2n + 1)2+ (2n2 + 2n)2= (2n2 + 2n + 1)2
(2n + 1)(2n + 1) + (2n2 + 2n)(2n2 + 2n) = (2n2 + 2n + 1)(2n2 + 2n + 1)
4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1
4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1
4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1
This proves that my ‘a’, ‘b’ and ‘c’ formulas are correct
Generalisations
Difference between each shortest side that is an odd integer. – 2
Difference between each middle side. – 2nd difference =4
Difference between each hypotenuse. – 2nd difference = 4
Difference between shortest and middle side. 2nd difference = 4
Difference between shortest and longest side. 2nd difference =5
Difference between middle and longest side. +1
2nd Difference is four
Difference between shortest and middle length sides generalisations.
1 7 17 31 49 71 97 127 161 199
6 10 14 18 22 26 30 34 38
2nd Difference is four
