Triangle 5
A2 + B2 = C2
112 + 602 = C2
121 + 3600 = C2
3721 = C2
√3721 = C2
61 = Side C
Again this is a Pythagoras triangle.
Triangle 6
A2 + B2 = C2
132 + 842 = C2
169 + 7056 = C2
7225 = C2
√3721 = C2
85 = Side C
I now can be sure that all the new triangles are Pythagoras.
Perimeter and Area
I am going to work out the perimeter and area of each triangle.
Perimeter
I will work out the perimeter by adding all the sides of the triangles and see what they total.
Triangle 1: 3 + 4 + 5 = 12
Triangle 2: 5 + 12 + 13 = 30
Triangle 3: 7 + 24 + 25 = 56
Triangle 4: 9 + 40 + 41 = 90
Triangle 5: 11 + 60 + 61 = 132
Triangle 6: 13 + 84 + 85 = 182
nth term for perimeter
12 30 56 90 132 182
\ / \ / \ / \ / \ /
18 26 34 42 50
\ / \ / \ / \ /
8 8 8 8
Now to work out the nth term I will use the formula explained on page 10.
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
12 + ( n – 1 ) 18 + ½ ( n – 1 ) ( n – 2 ) 8
12 + 18n – 18 + 4 [n2 – 3n + 2]
12 + 18n – 18 + 8n2 – 12n + 8
4n2 + 6n + 2
Area
To work out the area of the six triangles I will be using the formula:
Base x Height
2
Triangle 1: 4 x 3 = 6
2
Triangle 2: 12 x 5 = 30
2
Triangle 3: 24 x 7 = 84
2
Triangle 4: 40 x 9 = 180
2
Triangle 5: 60 x 11 = 330
2
Triangle 6: 84 x 13 = 546
2
nth term
Next, I am going to find out the nth term for the shortest side, the middle side and the hypotenuse side. I will do this because it makes it much easier to find out the length of a side for triangle n.
To work out the nth term I will use the formula: Tn = d n + ( a – d )
Where: a = the first term
d = common difference
n = term
Tn = nth term
nth term for shortest side
3 5 7 9 11 13 15
\ / \ / \ / \ / \ / \ /
2 2 2 2 2 2
Tn = d n + ( a – d )
= 2 n + ( 3 – 2 )
= 2 n + 1
I will now test this nth term to see if it is correct. I will do this by finding out the 7th term using the nth term then checking it against my sequence.
So when: n = 7
2 x 7 + ( 3 – 2)
14 + 1 = 15
This proves that the nth term is correct because it matches with what I worked out what would be the 7th term in my sequence.
nth term for middle side
4 12 24 40 60 84 112
\ / \ / \ / \ / \ / \ /
8 12 16 20 24 28
\ / \ / \ / \ / \ /
4 4 4 4 4
Because this has a second difference I will use a second formula to work out the nth term.
Forumla: Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
Where: Tn = nth term
A = the first term
D = the first number in the changing sequence
C = the second, continuous difference
N = term
I will now substitute the numbers from the sequence into the formula.
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
4 + ( n – 1 ) 8 + ½ ( n – 1 ) ( n – 2 ) 4
4 + 8n – 8 + 2 [n2 – 3n + 2]
4 + 8n – 8 + 2n2 – 6n + 4
2n2 + 2n + 0
I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.
So when n = 7
2 x 72 + 2 x 7 + 0
2 x 49 + 14
98 + 14 = 112
In my sequence under the 7th term I have 112. This proves that my nth term is correct.
nth term for hypotenuse side
I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.
5 13 25 41 61 85 113
\ / \ / \ / \ / \ / \ /
8 12 16 20 24 28
\ / \ / \ / \ / \ /
4 4 4 4 4
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
5 + ( n – 1 ) 8 + ½ ( n – 1 ) ( n – 2 ) 4
5 + 8n – 8 + 2 [n2 – 3n + 2]
5 + 8n – 8 + 2n2 – 6n + 4
2n2 + 2n + 1
I will now again test this nth term using the same method as above.
So when n = 7
2 x 72 + 2 x 7 + 1
2 x 49 + 14 + 1
98 + 14 + 1 = 113
Both say the 7th term is 113, so it must be correct.
Testing the nth terms
I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.
nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2
So: ( 2n + 1)2 + ( 2n2 + 2n )2 = (2n2 + 2n + 1)2
I will now expand the brackets for each nth term.
( 2n + 1 )2 expanded:
( 2n + 1) ( 2n + 1)
= 4n2 + 4n + 1
( 2n2 + 2n )2 expanded:
( 2n2 + 2n ) ( 2n2 + 2n )
= 4n4 + 4n3 + 4n3 + 4n2
= 4n4 + 8n3 + 4n2
(2n2 + 2n + 1)2 expanded:
(2n2 + 2n + 1) (2n2 + 2n + 1)
= 4n4 + 4n3 + 2n2 + 4n3 + 4n2 + 2n +2n2 + 2n + 1
= 4n4 + 8n3 + 8n2 + 4n + 1
So: 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 = 4n4 + 8n3 + 8n2 + 4n + 1
8n2 + 4n + 1 + 8n3 + 4n4 = 4n4 + 8n3 + 8n2 + 4n + 1
4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1
This proves that the nth terms are all right and work with Pythagoras theorem.
CONCLUSION
I have now generalized the problem given to me at the beginning. I do not think I can go any further with it. I have proved and tested all my formulas.
However, I do think I could extent this though by looking at a different family of Pythagorean triplets. In this family there was a difference of 1 between the values of the middle side and longest side. I will now look at a family that has a difference of 2. I want to see if there if the nth terms will be linked or have something in common.
I can now produce any triplet in this family with my nth terms easily.
EXTENSION
In this extension I will do the same with what I did with the first family but with the new family and sides.
To work out the new family I will double all the values from the last Pythagorean triplets.
So Triangle 1 will be:
3, 4, 5 doubled to 6, 8, 10
Triangle 2
5, 12, 13 doubled to 10, 24, 26
Triangle 3
7, 24, 25 doubled to 14, 48, 50
Here is the new family:
I will now check to see if these triangles are Pythagorean triplets by putting the shortest side and middle find into the formula and see if it gives me the answer for the hypotenuse.
Triangle 1
A2 + B2 = C2
62 + 82 = C2
36 + 64 = C2
100 = C2
C = 10
This triangle is a member of a Pythagorean triplet.
Triangle 2
A2 + B2 = C2
102 + 242 = C2
100 + 576 = C2
676 = C2
C = 26
Again, this triangle is a member of a Pythagorean triplet.
Triangle 3
A2 + B2 = C2
142 + 482 = C2
196 + 2304 = C2
2500 = C2
C = 50
All these triangles are Pythagorean triplets.
Like last time I will now put these triangle into a table for the same reasons.
Everything that I add to my table after this point will be in blue.
I will now work out what the next three triangles in the family will be.
Sequence for shortest side
6 10 14 18 22 26
\ / \ / \ / \ / \ /
4 4 4 4 4
Every time 4 is being added to the previous number. I can work out that the next numbers will be 18, 22, and then 26.
Sequence for middle side
8 24 48 80 120 168
\ / \ / \ / \ / \ /
16 24 32 40 48
\ / \ / \ / \ /
8 8 8 8
Here there is a continuous second difference of 8. I can tell that the number after 48 will be 80. I worked this out by finding the difference between 24 and 48 (24), adding 8 to it (32) then adding it on to 48.
Sequence for hypotenuse side
10 26 50 82 122 170
\ / \ / \ / \ / \ /
16 24 32 40 48
\ / \ / \ / \ /
8 8 8 8
On this sequence the second difference is 8, and by adding 8 every time to the first non continuous difference I can tell that the next numbers will be 82, 122 then 170.
Test that the three new triangles are in the Pythagorean family
I will again, like I did with the last family, test to see if the three new triangles I have got numbers for do comply with Pythagoras theorem. I will to this by adding side a2 and side b2 and see if I get the answer I get matches what I got for side C in my table.
Triangle 4
A2 + B2 = C2
182 + 802 = C2
324 + 6400 = C2
3360 = C2
√6724 = C2
82 = Side C
This answer matches with what I predicted with my table. Therefore it is defiantly a Pythagoras triangle.
Triangle 5
A2 + B2 = C2
222 + 1202 = C2
484 + 14400 = C2
14884 = C2
√14884 = C2
122 = Side C
Again this is a Pythagoras triangle.
Triangle 6
A2 + B2 = C2
262 + 1682 = C2
676 + 28224 = C2
28900 = C2
√28900 = C2
170 = Side C
I now can be sure that all the new triangles are Pythagoras.
nth term of new family
Next, like last time, I am going to find out the nth term for the shortest side, the middle side and the hypotenuse side. I will do this because it makes it much easier to find out the length of a side for triangle n. I will use the same formula I describe when investigating the last family.
nth term for shortest side
6 10 14 18 22 26 30
\ / \ / \ / \ / \ / \ /
4 4 4 4 4 4
Tn = d n + ( a – d )
= 4 n + ( 6 – 4 )
= 4 n + 2
I will now test this nth term to see if it is correct. I will do this by finding out the 7th term using the nth term then checking it against my sequence.
So when: n = 7
4 x 7 + ( 6 – 4)
28 + 2 = 30
This proves that the nth term is correct.
nth term for middle side
8 24 48 80 120 168 224
\ / \ / \ / \ / \ / \ /
16 24 32 40 48 56
\ / \ / \ / \ / \ /
8 8 8 8 8
Because this has a second difference I will use a second formula to work out the nth term that I have noted above when working out the nth term for the middle side on the first family.
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
8 + ( n – 1 ) 16 + ½ ( n – 1 ) ( n – 2 ) 8
8 + 16n – 16 + 4 [n2 – 3n + 2]
8 + 16n – 16 + 4n2 – 12n + 8
4n2 + 4n + 0
I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.
So when n = 7
4 x 72 + 4 x 7 + 0
4 x 49 + 28
196 + 28 = 224
In my sequence under the 7th term I have 224. This proves that my nth term is correct.
nth term for hypotenuse side
I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.
10 26 50 82 122 170 226
\ / \ / \ / \ / \ / \ /
16 24 32 40 48 56
\ / \ / \ / \ / \ /
8 8 8 8 8
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
10 + ( n – 1 ) 16 + ½ ( n – 1 ) ( n – 2 ) 8
10 + 16n – 16 + 4 [n2 – 3n + 2]
10 + 16n – 16 + 4n2 – 12n + 8
4n2 + 4n + 2
I will now again test this nth term using the same method as above.
So when n = 7
4 x 72 + 4 x 7 + 2
4 x 49 + 28 + 2
196 + 28 + 1 = 226
Both say the 7th term is 226, so it must be correct.
Testing the nth terms of new family
I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.
nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2
So: ( 4n + 2)2 + ( 4n2 + 4n )2 = (4n2 + 4n + 2)2
I will now expand the brackets for each nth term.
( 4n + 2 )2 expanded:
( 4n + 2) ( 4n + 2)
= 16n2 + 16n + 4
( 4n2 + 4n )2 expanded:
( 4n2 + 4n ) ( 4n2 + 4n )
= 16n4 + 16n3 + 16n3 + 16n2
= 16n4 + 32n3 + 16n2
(4n2 + 4n + 2)2 expanded:
(4n2 + 4n + 2) (4n2 + 4n + 2)
= 16n4 + 16n3 + 8n2 + 16n3 + 16n2 + 8n +8n2 + 8n + 4
= 16n4 + 32n3 + 32n2 + 16n + 4
So:
16n2 + 16n + 4 + 16n4 + 32n3 + 16n2 = 16n4 + 32n3 + 32n2 + 16n + 4
32n2 + 16n + 4 + 32n3 + 16n4 = 16n4 + 32n3 + 32n2 + 16n + 4
16n4 + 32n3 + 32n2 + 16n + 4 = 16n4 + 32n3 + 32n2 + 16n + 4
This proves that the nth terms are all right and work with Pythagoras theorem.
Extension conclusion for ‘Difference of 2’
Comparing the 2 nth terms I can conclude that when there is a different of 2 between the middle side and the hypotenuse side the nth term is doubled from when there was a difference of 1. I predict that when there is a difference of 3 between the middle side and the hypotenuse side and nth term will be tripled.
Difference of 3
I will now look at a family with a difference of 3 to see if my prediction will be correct.
To get these triangle sides I have just trebled the numbers from the first family. To make sure that these are correct I will pick a triangle and check whether it is a Pythagorean triplet.
Triangle 4 = A2 + B2 = C2
272 + 1202 = C2
726 + 14,400 = C2
15,126 = C2
123 = C
This proves that this triangle and all of the new ones are Pythagorean triplets of a new family.
nth term of the new family
I am now going to find out the new nth terns for all the sides using the same techniques as I have used before and for the same reasons I have stated above.
nth term for shortest side
9 15 21 27 33 39 45
\ / \ / \ / \ / \ / \ /
6 6 6 6 6 6
Tn = d n + ( a – d )
= 6 n + ( 9 – 6 )
= 6 n + 3
I will now like last time test to see if this nth term is correct by finding out the 7th term using the nth tern then checking it against my sequence.
So when: n = 7
6 x 7 + ( 9 – 6 )
42 + 3 = 45
This proves that the nth term is correct.
nth term for middle side
12 36 72 120 180 252 336
\ / \ / \ / \ / \ / \ /
24 36 48 60 72 84
\ / \ / \ / \ / \ /
12 12 12 12 12
Because this has a second difference I will use a second formula to work out the nth term that I have noted above when working out the nth term for the middle side on the first family.
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
12 + ( n – 1 ) 24 + ½ ( n – 1 ) ( n – 2 ) 12
12 + 24n – 24+ 6 [n2 – 3n + 2]
12 + 24n – 24 + 6n2 – 18n + 12
6n2 + 6n + 0
I will now test this nth term again like last time to see if it is correct. I will find out the 7th term by using the nth term I have just worked out then check it against what I have in my sequence.
So when n = 7
6 x 72 + 6 x 7 – 0
6 x 49 + 42
294 + 42 = 336
In my sequence under the 7th term I have 336. This proves that my nth term is correct.
nth term for hypotenuse side
15 39 75 123 183 255 339
\ / \ / \ / \ / \ / \ /
24 36 48 60 72 84
\ / \ / \ / \ / \ /
12 12 12 12 12
I will find out this nth term using the same formula I used finding out the middle side, as it has a second difference.
Tn = a + ( n – 1 ) d + ½ ( n – 1 ) ( n – 2 ) c
15 + ( n – 1 ) 24 + ½ ( n – 1 ) ( n – 2 ) 12
15 + 24n – 24+ 6 [n2 – 3n + 2]
15 + 24n – 24 + 6n2 – 18n + 12
6n2 + 6n + 3
I will now again test this nth term using the same method as above.
So when n = 7
6 x 72 + 6 x 7 + 3
6 x 49 + 45
294 + 45 = 339
In my sequence under the 7th term I have 336. This proves that my nth term is correct.
Both say the 7th term is 336, so it must be correct.
Testing the nth terms of ‘difference of 3’
I am now going test all the nth terms. I will do this by seeing if they comply with Pythagoras theorem.
nth term for shortest side2 + nth term for middle side2 = nth term for hypotenuse side2
So: ( 6n + 3)2 + ( 6n2 + 6n )2 = (6n2 + 6n + 3)2
I will now expand the brackets for each nth term.
( 6n + 3 )2 expanded:
( 6n + 3) ( 6n + 3)
= 36n2 + 36n + 9
( 6n2 +6 )2 expanded:
( 6n2 + 6n ) ( 6n2 + 6n )
= 36n4 + 36n3 + 36n3 + 36n2
= 36n4 + 72n3 + 36n2
(6n2 + 6n + 3)2 expanded:
(6n2 + 6n + 3) (6n2 + 6n + 3)
= 36n4 + 36n3 + 18n2 + 36n3 + 36n2 + 18n +18n2 + 18n + 9
= 36n4 + 72n3 + 72n2 + 36n + 9
So:
36n2 + 36n + 9 + 36n4 + 72n3 + 36n2 = 36n4 + 72n3 + 72n2 + 36n + 9
72n2 + 36n + 9 + 72n3 + 36n4 = 36n4 + 72n3 + 72n2 + 36n + 9
36n4 + 72n3 + 72n2 + 36n + 9 = 36n4 + 72n3 + 72n2 + 36n + 9
This proves that the nth terms are all right and work with Pythagoras theorem.
Extension conclusion for ‘Difference of 3’ and final conclusion
Comparing the 3 nth terms I can conclude that when there is a different of 2 between the middle side and the hypotenuse side the nth term is doubled from when there was a difference of 1. When there is a different of 3 between the middle side and the hypotenuse side the nth term is tripled. I predict that when there is a difference of 4 between the middle side and the hypotenuse side and nth term will be quadrupled and so on.