I can test this by working out 52, 122 and 132,
And if I add 25 to 144, it does equal 169, so ‘a’ will go up by two each time, and ‘c’ is always equal to ‘b + 1’ when ‘a’ is an odd number.
Now I can start to try and work out the nth terms for a, b, c, the perimeter and the area. I will start by working out the nth term for a.
Here, I have written out the first 5 numbers in the sequence of ‘a’, and written out the differences between them. I have also written what the previous number in the sequence would be.
To make the nth term for ‘a’, I take the common difference, 2, and I multiply it by ‘n’. Then, because the previous number to 3 would be ‘+1’, I simply add this to my ‘2n’ to make the nth term for ‘a’, ‘2n + 1’. Before coming to the conclusion that ‘2n + 1’ is correct I will be checking it.
This formula for ‘a’ is correct as 2n + 1 does equal ‘a’.
a = 2n + 1
The next formula I want to find is the nth term for ‘b’. I will use the same method as I used to find the nth term for ‘a’. I have written out the first 5 numbers in the sequence of ‘b’.
This sequence is different to the sequence for ‘a’ in that there is not a common difference until the second line. When this happens, you half the common difference, and multiply it by ‘n2’ so, in this case, I would get ‘2n2’. Then I will substitute ‘n’ for each of the first 5 numbers, and work out what it would give. I will call this new sequence ‘sequence 2’. Next, you compare it to the original sequence. So I will work out ‘2n2’ for the first 5 numbers, and compare it to the original sequence for ‘b’.
Here I have substituted ‘n’ for the numbers from 1 to 5, and found ‘sequence B’. Now I will compare it with the original sequence for ‘b’.
I have worked out the differences between the ‘sequence 2’ and the original sequence. The differences will form a new sequence, which I will call ‘sequence 3’.
Now, I work out the nth term for this sequence. There is a common difference on the first line, so I just multiply 2 by ‘n’, and because the previous number would be 0, I don’t need to add or subtract anything, so the nth term for this sequence is simply ‘2n’.
All I have to do now is to add the two formulas, ‘2n2’ and ‘2n’, and then I will have the nth term for ‘b’-‘2n2 + 2n’. Like I did with the nth term for ‘a’, I will test this formula by substituting ‘n’ for 1.
I changed ‘n’ in the equation b = 2n2 + 2n to 1 and worked out the answer
This answer is correct because the first number in the first term for ‘b’ is 4.
This formula for ‘b’ is correct as 2n2 + 2n does equal ‘b’.
b= 2n2 + 2n
The third nth term that I want to find out is the nth term for ‘c’. This is easier for me to find, because I already know that ‘c’ is always ‘b + 1’ when ‘a’ is an odd number, and I know the formula for ‘b’ which is 2n2 + 2n. I just have to take the nth term for ‘b’, and add 1 at the end.
b = 2n2 + 2n
c = 2n2 + 2n + 1
Like I did with the nth term for ‘a’ and ‘b’, I will test this formula by substituting ‘n’ for 1.
I changed ‘n’ in the equation c = 2n2 + 2n + 1 to 1 and worked out the answer
This answer is correct because the first number in the first term for ‘c’ is 5.
This formula for ‘c’ is correct as 2n2 + 2n + 1 does equal ‘c’.
c = 2n2 + 2n + 1
Perimeter
To find the perimeter you have to add a, b and c together.
If ‘a’ is 3, ‘b’ is 4 and ‘c’ is 5, like the 1st term, the perimeter will be 12 because 3+4+5=12.
If ‘a’ is 5, ‘b’ is 12 and ‘c’ is 13, like the 2nd term, the perimeter will be 30 because 5+12+13=30.
The formula to find the perimeter is correct as a + b + c does equal the perimeter
a + b + c
Area
To find the area you have to multiply ‘a’ and ‘b’ together and then divide the answer by 2.
If ‘a’ is 3 and ‘b’ is 4, like the 1st term, the area will be 6 because 3x4=12 and 12÷2= 6
If ‘a’ is 5 and ‘b’ is 12, like the 2nd term, the area will be 30 because 5x12= 60 and 60÷2= 30
The formula to find the area is correct as (a x b)/2 does equal the area
(a x b)/2
I have worked out the nth terms for ‘a’, ‘b’ and ‘c’ and I have proved that they are correct, I will try to do the same with Pythagorean triples where ‘a’ is an even number.
Evens
Again, I will draw up a table containing the first eight Pythagorean triples with ‘a’ being an even number, and shows the lengths of the sides, their perimeters and their areas. Then I will be able to see if there are any connections or relationships between the numbers, and I will be able to find the nth term for each side, area and perimeter.
The first nth term that I want to find from this table is the nth term for ‘a’. I will do exactly the same things as I did to find the nth term for the odd ‘a’ sequence, but the numbers will be different.
I take the ‘2n’ and the ‘+ 4’, and put them together to make my formula for the even ‘a’ sequence, ‘2n + 4’. I now need to check this in the same way I did for all the formulas in the odd ‘a’ sequence, by substituting ‘n’ for 1.
a = 2 x 1 + 4
a = 1 + 4
a = 6
This formula for ‘a’ is correct as 2n + 4 does equal ‘a’.
a = 2n + 4
Now I will try and find the nth term for ‘b’ when ‘a’ is even. Here are the first five numbers in the sequence, and the differences between them, until there is a common difference.
Now, I apply this ‘n2’ to the numbers 1, 2, 3, 4, and 5, to give me a new sequence, which I will call ‘sequence 2’, and compare it with the original sequence for ‘b’, to give me another new sequence, ‘sequence 3’. Then, I will work out the nth term for ‘sequence 3’, and put ‘n2’ before it to make the nth term for ‘b’ when ‘a’ is even.
4n + 3
b = n2 + 4n + 3
I take the 4n + 3 and add it to make the formula for the even ‘b’ sequence, ‘n2 + 4n + 3. I now need to check this by substituting ‘n’ for 1.
b = n2 + 4n + 3
b = 12 + 4 x 1 + 3
b = 1 x 1 + 4 x 1 + 3
b = 1 + 4 + 3
b = 8
This formula for ‘b’ is correct as n2 + 4n + 3 does equal ‘b’.
b = n2 + 4n + 3
The next nth term that I need to find is for ‘c’ when ‘a’ is even. I have already found out that when ‘a’ is an even number, ‘b’ is always two more than ‘c’, and I can use this to find the nth term for ‘c’. The nth term for ‘b’ is ‘n2 + 4n + 3’, and ‘c’ is 2 more than this, then the nth term for ‘c’ must be n2 + 4n + 5. I will test my prediction using the number 1, which will replace ‘n’.
b = n2 + 4n + 5
b = 12 + 4 x 1 + 5
b = 1 x 1 + 4 x 1 + 5
b = 1 + 4 + 5
b = 10
This formula for ‘c’ is correct as 2n + 4 does equal ‘a’.
b = n2 + 4n + 3
Perimeter
To find the perimeter you have to add a, b and c together.
If ‘a’ is 6, ‘b’ is 8 and ‘c’ is 10, like the 1st term, the perimeter will be 24 because 6+8+10=24.
If ‘a’ is 8, ‘b’ is 15 and ‘c’ is 17, like the 2nd term, the perimeter will be 40 because 8+15+17=40.
The formula to find the perimeter is correct as a + b + c does equal the perimeter
a + b + c
Area
To find the area you have to multiply ‘a’ and ‘b’ together and then divide the answer by 2.
If ‘a’ is 6 and ‘b’ is 8, like the 1st term, the area will be 24 because 6x8=48 and 48÷2= 24
If ‘a’ is 8 and ‘b’ is 15, like the 2nd term, the area will be 60 because 8x15= 120 and 120÷2= 60
The formula to find the area is correct as (a x b)/2 does equal the area
(a x b)/2
I have worked out the nth terms for ‘a’, ‘b’ and ‘c’ and I have proved that they are correct
Conclusion table
