3 5 7 9 11
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.
Let’s see…
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. To see if I’m correct I will now test this formula.
2n+1
2x1+1=3 (correct)
Just in case I will test this formula in the next term:
2n+1
2x2=4
4+1=5 (correct)
I now have to work out the formula for the middle side. I predict that the formula has got to do with something about the differences in lengths. (For instance in this case is 4), most likely 4n.However, because 4 is the difference, the formula must be n². I now believe that the answer will have something to do with 4n². So, I will now write the answers for 4n².
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4n² work for the first term, but, it then collapses after this, as the difference between 4n² gets larger, the thing you notice is that the difference in the 2nd term between 4n² and the middle side is the middle side for the term before. This goes for all the other terms from the 2nd.
This means that if I subtract the previous term, then I should in theory get the correct answer.
16-4 = 12
36-12 = 24
64-24 = 40
Etc.
So, the equation I have so far is:
4n²- (previous middle side) = middle side
All the previous terms is, (n - 1), so if I put this into the above formula, then it should give me my middle side.
4n² - 4(1-1) ² = middle side
This should in theory give me my middle side. I will test my theory with the first term. wweb ebw stebebud eeb ebnt ceb enebtral ebcoeb uk;
4 x 12 - 4(1-1) = 4
4 x 1 – 4 x 0² = 4
4 – 4 x 0 = 4
4 – 0 = 4
4 = 4
My formula works for the first term. I will now check if this formula works for the next term.
4 x 2² - 4(2 – 1) ² = 12
4 x 4 – 4 x 1² = 12
16 – 4 x 1 = 12
16 – 4 = 12
12 = 12
My formula also works for the 2nd term. You may be thinking that this is the correct formula but just to check, I will check this formula for the 3rd term.
4 x 3² - 4(3-1) ² = 24
4 x 4 – 4 x 2² = 24
36 – 4 x 4 = 24
36 – 16 = 24
20 = 24
My formula did not work for the 3rd term. It now looks as if “4n² - 4(n – 1) ² ” is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.
4 x 4² - 4(4 – 1) ² = 40
4 x 16 – 4 x 3² = 40
64 – 36 = 40 wwdb dbw stdbdbud edb dbnt cdb endbtral dbcodb uk.
28 = 40
My formula doesn’t work for the 4th term either. I can now safely say that
4n² - 4(n – 1) ² is definitely not the correct formula for the middle side.
I believe the problem with 4n² - 4(n – 1) was that 4n², once you start using the larger numbers, it becomes far too high to bring it back down to the number I want for the middle side. Also, 4(n – 1) ² is not as small when it gets larger so it doesn’t bring the 4n² down enough, to equal the middle side.
I will now look at the difference to see if I can find a pattern there.
1 4 11 20 31
3 7 9 11
2 2 2
The difference here is 2, which means that the answer will involve 2 and n². I will try 2n². I can see that the difference between 2n² and the middle number is the 2 times table. The 2 times table in the nth term is 2n. I now think 2n² + 2n is the correct formula. 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. To start with, I am going to draw out a table containing the middle and longest sides. from www.studentcentral.co.uk
From this table I know that there is only one difference, which is +1 between the middle and longest side. So:
(Middle side) + 1 = longest side
I predict that this formula will be the correct one for the longest side:
2n² + 2n + 1 = longest side. wwbg bgw stbgbgud ebg bgnt cbg enbgtral bgcobg uk.
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 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk!
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 out 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
So 12 is the perimeter for the first term
2nd term 5 + 12 + 13 = 30
3rd term 7 + 24 + 25 = 56
And so on. All I have to do is put all the 3 formulas together.
Perimeter = (shortest side) + (middle side) + (longest side) f0KAU1K from f0KAU1K student f0KAU1K central f0KAU1K co f0KAU1K uk
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 6n + 2
If I have done my calculations properly then I should have the right answer. To check this I am going use the 4th, 5th and 6th terms.
4th term:
4n² + 6n² + 2 = perimeter
4 x 4² + 6 x 4 + 2 = 9 + 40 + 41
64 + 24 + 2 = 90
90 = 90
It works for the 4th term
Let’s see if it works for the 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 wwed edw stededud eed ednt ced enedtral edcoed uk!
Like the area 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)