The numbers 5,12,13 satisfy the conditiona2 + b2 = c252 + 122 = 13252 = 5 x 5 = 25122 = 12 x 12 = 144132 = 13 x 13 = 169and so 52+ 122 = 25 + 144 = 169 = 132b) The numbers 7, 24, 25a2 + b2 = c272 + 242 = 25because 72 = 7 x7 = 49242 = 24 x 24 = 576252 = 25 x 25 = 625and so 72 + 242 = 49 + 576 = 625 = 25 2The numbers above satisfy similar condition that is pythagors theorem. I have tested the numbers by putting them into the formula a2 + b2 = c2. This is means the numbers are a Pythagorean triple because they satisfy the condition a2 + b 2 = c22)Length of shortest sideLength of middle sideLength of longest sidePerimeterArea3451265121330307242556849404190180a) I found the perimeter for the sequence 5, 12, 13 and 7, 24, 2, by simply finding the sum (add) all the lengths of the three sides together to get the perimeter.

E.g. Perimeter = shortest +middle + longest length5 + 12 + 13 = 30 units7 + 24 + 25 = 56 unitsI found the area for the sequences 5, 12, 13 and 7, 24, 25, by simply finding the product (multiply) the shortest and middle and halving the answer.E.g. Area = 1/2 x shortest x middle length1/2 x 5 x 12 = 30 square units1/2 x 7x 24 = 84 square unitsI used this method for the area because it is a right-angled triangle. The dotted lines above show that it is actually half a square.

Since the area of a square is length x width. Therefor the area would be the same equation divided by 2.Patternsb) I found the next result in the sequence by finding the pattern that each side had.

When I knew the pattern for each side I could easily calculate the area and perimeter using the equations above.Length of Shortest side> The sequence 3, 4, 5 are going up in two’s i.e. the difference between the next sequence and the previous one is 2.3 + 2 = 54 + 2 = 65 + 2 = 77 + 2 = 9 The number in the sequenceFrom this I can form a formula which is Un = Un-1 + 2. This formula shows that if you add 2 to the previous sequence then the sum will be the next sequence. This is shown by the equations above.

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With this formula the next sequence can ONLY be know if the previous one is know.; I drew a table to present my findings clearly.nUn1st difference13225237249Length of Middle side; The sequence was 4, 12, and 24. To find the next sequence I drew a table as there was more than one difference.nUn1st difference2nd difference148212412324416440The table shows that the first differences in the table are not the same therefor, I then worked out the second difference. I found the second differences to be constant – all 4This pattern enabled me to calculate the fourth term (highlighted in blue). I added 4 to the first difference in series (4 + 12 = 16), the sum was then added to Un in series (16 + 24 = 40).; I extended the table to see if I cold notice any other more reliable patterns.

nUn2n(2n + n) x n142421281232418244403240Another pattern I noticed was that if you multiplied by a whole number Un was found, by doing this I further noticed that the difference between the multiples was 2.E.g.if 1 x 4 = 42 x 6 = 83 x 8 = 24then 4 x 10 = 40; From this information I could form a formula by finding the number that you multiply n with to get Un. So I formed the formula 2n +n, that would equal the number that you multiply.

Once I found this I then knew that by mutilplying it by n I could get Un.E.g. (2n +n) x n = Un(2 x 2 + 2) x 2 = 12(2 x 3 + 2) x 3 = 24(2 x 4 + 2) x 4 = 40Length of the Longest side; The sequence was 5, 13, 25. To find the next number in the sequence I used the same method as I did to find the middle length as they both followed similar patternsnUn1st difference2nd difference158213412325416441The table shows that the first differences in the table again are not the same, therefor I then worked out the second difference. I found the second differences to be constant – all 4.

This pattern enabled me to calculate the fourth term (highlighted in blue). Iadded 4 to the first difference in series (4 + 12 = 16), the sum was then added to Un in series (16 + 24 = 40).; Another pattern I noticed was that if you added 1 to the middle length of the same sequence then the longest length was obtained.E.

g.if 4 + 1 = 512 + 1 = 1324 + 1 = 25then 40 + 1 = 41I further checked that the 4th term was correct by checking that the sequences satisfied a similar condition of ( smallest number )2 = ( middle number) 2 = (largest number) 2E.g.if 32 + 42 = 52then 92 + 242 = 90 2because 92 + 242 = 81 + 576 = 657 = 412Hence this means that all the equations are pythagrean triples because they all satisfy the condition a2 + b2 = c2PerimeterI found the perimeter after I had worked out all three lengths. I then added all the sides of the sides to get the perimeter. The equation for the perimeter is shortest + middle + longest length. So I used this to find the perimeter.E.

g.If 3 + 4 + 5 = 12 units5 + 12 +13 = 30 units7 + 24 +25 = 56 unitsthen 9 + 40 + 41 = 90 unitsAreaI found the area by using the equation Area = 1/2 ( length x width). As it is a Pythagorean triple I knew that these lengths would form a right-angled triage.E.g.If 1/2 (3 x 4 ) = 51/2 (5 x 12) = 301/2 x (7 x 24) = 84then 1/2 x ( 9 x 40) = 90I extended the table using lengths of positive integer. I started by looking at triangles which the shortest side is an odd number of units in length. This would enable me to work out the sequence in relation to which term it is.

PredictionsFrom the patterns I have previously noted I predict that the following four terms are correct.Nth termLength of shortest sideLength of middle sideLength of longest sidePerimeterArea13451262512133030372425568449404190180511606113233061384851825467151121132408408161441453061152Shortest lengthI predict that the 5th term for the shortest length will be 11 because using the patterns identified I know that if you add 2 to the previous term it will equal the next term.E.g. 9 = 2 = 11. I also predict that a will be 2 because I notice that the difference is 2 also.I will test that this prediction is correct I will work out the formula. It is a linear equation because the difference between any two consecutive terms is the same.

Length- linear equation, tn =an + b.Finding the ruleSo I will be finding a rule for the sequence 3, 5, 7, 9…nUn1st difference13225237249The table above shows that the difference between any two consecutive terms is 2.The rule is tn = 2n + at1 = 2 x 1 + a ( replacing n by 1 )3 = 2 + a (since t1 =3)1 = a (subtracting 2from both sides)Hence the rule is tn = 2n + 1; I noticed a connection between the rule for the sequences and the row of 2’s. This is that the rule has 2 in it and the second row of difference is also 2.Testing & provingTo test the formula is correct will write down the first eight sequences given by the rule.

tn = 2n + 1> Replace n by 1, 2, 3, 4, 5, 6, 7, 8t1 = 2 x 1 + 1 t2 = 2 x 2 + 2 t3 = 2 x 3 + 3 t4 = 2 x 4 + 4= 3 = 5 = 7 = 9t5 = 2 x 5 + 5 t6 = 2 x 6 + 6 t7 = 2 x 7 + 7 t8 = 2 x 8 + 8= 11 = 13 = 15 = 17> I will also test the formula by using the rule to find the 18th term of the sequence.Using tn = 2n + 1, t18 = 2 x 18 + 1= 37> I can even use the formula in an algebra equation to find the term of a sequence 18.Using tn = 2n + 1, 18 = 2n + 1I can solve the equation 18 = 2n + 1 to find n. using the balance method the solution is: 37 = 2n + 136 = 2n (subtracting 1 to each side)18 = n (dividing both sides 2)Therefor the t18 = 37. That is the 18th term is 37.Overall my results proved to agrere with my original prediction as I used a linear equation which tested with other methods was correct.PredictionUsing the patterns found earlier I predict that the four terms in the table 9pg5 0 are correct.

Middle lengthnUn1st difference2nd difference148212412324416440I predict that the 5th term will be 60 because the difference between the difference of the previous term is 4. I therefor know that it is a quadratic equation because the difference between the difference of any two consecutive terms is the same. The difference method below enables me to predict that the highest power of n in a rule for a sequence is n2 because it goes to the second difference. If the sequence were to go to the third difference then it would be n3.A quadratic rule for a sequence is tn = an2 = bn + c. The following equations will aid me in my investigation to make another prediction. There is a connection between the second row of difference and a.

tn = 2n2 + 3 tn = 3n2 + 5nThere is a relationship between the second row of difference and a. This is that the second row of difference halved is equal to a. Therefor I predict that if the second row of difference is 4 then a will be equal to 2.

Finding the ruleI will investigate further to find b and cI can then use this to find the quadratic rule for the sequence 4, 12, 24, 40….I will investigate further to find ba and c.

For tn = an2 = bn + c.4 12 24 408 12 164 42a = 4 3a + b = 8 a + b + c = 4a = 2 3 x 2 + b = 8 2 + 2 + c = 46 + b = 8 4 + c = 4b = 8 – 6 c = 4 – 4b = 2 c = 0The rule is 2n2 + 2nTesting and proving> To test the formula is correct I will write down the first eight sequences given by the rule. I can then compare them with those I got using the patterns I noticed.

Using tn = 2n2 + 2n.t1 = 2 x 12 + 2 x 1 t2 = 2 x 22 + 2 x 2 t3 = 2 x 32 + 2 x 34 12 24t4 = 2 x 42 + 2 x 4 t5= 2 x 52 + 2 x 5 t6= 2 x 62 + 2 x 640 60 84> I will also test the formulae by using the rule to find the 8th term of the sequenceUsing tn = 2n2 + 2n, t8 = 2 x 82 + 2 x 8= 144> I can also use the formula in algebra equation to find the term of the sequence 8.