Applying this formula you get: “2+(n-1)3+1/2(n-1)(n-2)1” simplified you get “n +3n”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
1 +3x1=4 2 +3x2=12
This proves that my formula is correct.
Longest side
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a second difference, four between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “5+(n-1)x8+1/2x(n-1)(n-2)4” simplified you get “2n +2n+1”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
2x122x1+1=5 2x22+2x2+1=13
This proves that my formula is correct.
Perimeter
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a second difference, eight between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “12+(n+1)18+1/2(n-2)(n-2)x8” simplified you get “4n2+6n+2”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
4+12+6x1+2=12 4x22+6x1+2=30
This proves that my formula is correct.
Area
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a third difference, twelve between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “6+(n-1)24+1/2(n-1(n-2)12” simplified you get “6n2+6n-6”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
6x12+6x1-6=6 6x22+6x2-6=30
This proves that my formula is correct.
Now I will generalize everything. I have investigated some more Pythagoras triples and found out that the Pythagorean triple 9,12,15 does not satisfy my favourite triples that are even. I have looked on the internet and found a list of Pythagorean triples and found that 9,12,15 is a multiple of the famous 3,4,5 Pythagorean triple. Multiplying this Pythagorean triple will also give you other Pythagorean triples.
X2 =6,8,10
X3 =9,12,15
X4 =12,16,20
X5 =15,20,25
I am now going to investigate even pythagoras triples.
A Pythagorean Triple is a triple of natural numbers (a,b,c) such that: “c2 = a2 + b2.
It can be shown (this is a good exercise!) that all pythagorean triples can be written as: “a = x2 - y2, b = 2xy, c = x2 + y2” where x > y > 0 are natural numbers.
Following is a list of all pythagorean triples where a, b, c are all no larger than 100.
(3,4,5), (5,12,13), (6,8,10), (7,24,25), (8,15,17), (9,12,15), (9,40,41), (10,24,26), (11,60,61), (12,16,20), (12,35,37), (13,84,85), (14,48,50), (15,20,25), (15,36,39), (16,30,34), (16,63,65), (18,24,30), (18,80,82), (20,21,29), (20,48,52), (21,28,35), (21,72,75), (24,32,40), (24,45,51), (24,70,74), (25,60,65), (27,36,45), (28,45,53), (28,96,100), (30,40,50), (30,72,78), (32,60,68), (33,44,55), (33,56,65), (35,84,91), (36,48,60), (36,77,85), (39,52,65), (39,80,89), (40,42,58), (40,75,85), (42,56,70), (45,60,75), (48,55,73), (48,64,80), (51,68,85), (54,72,90), (57,76,95), (60,63,87), (60,80,100), (65,72,97).
I filled in the rest of the table by using my research.
Shortest side
I have worked out the nth term formula for the shortest side. This is how I did it:
The common difference between each term is two. From my class work I know I can work out the nth term formula using this equation:
“dn+(a-d)”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the common difference (in this case 2)
Applying this formula you get: “2n+(3-2)”, simplified you get: “2n+4”
To test this formula I will apply the formula to the 1st term and 2nd term.
1st term: 2nd term:
2x1+4=6 2x2+4=8
This proves that my formula is correct.
Middle side
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a second difference, four between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “8+(n-1)7+1/2(n-1)(n-2)2” simplified you get “n2+4n+3”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
12+4x1+3=8 22+4x2+3=
This proves that my formula is correct.
Longest side
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a second difference, four between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “10+(n-1)7+1/2(n-1)(n-2)7” simplified you get “n2+4n+5”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
12+4x1+5=10 12+4x1+5=17
This proves that my formula is correct.
Perimeter
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a second difference, eight between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “24+(n+1)16+1/2(n-2)(n-2)4” simplified you get “2n2+10n+12”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
2x12+10x1+12=24 2x2+ 10x2+12=40
This proves that my formula is correct.
Area
I have worked out the nth term formula for the middle side. This is how I did it:
This sequence has a third difference, twelve between each term. From my class work I know I can work out the nth term formula using this equation:
“a+(n-1)d+1/2(n-1)(n-2)c”.
- “a” is the value of the first term in the sequence.
- “d” is the value of the first difference.
- “c” is the value of the change in difference between one difference and the next.
Applying this formula you get: “624+(n-1)36+1/2(n-1(n-2)6” simplified you get “3n +27n-6”
To test this formula I will apply the formula to the 1st and 2nd terms.
1st term: 2nd term:
3x1 +27x1-6=6 3x2 +27x2-6=30
This proves that my formula is correct.
