This method is not part of the IGCSE syllabus but it's useful to know:"find a formula for the sequence 5, 14, 29, 50, 77, ... Since you know that the sequence is quadratic, you are looking for an expression an^2 + bn + c for the n-th term. With n = 1 we have a + b + c = 5, with n = 2 we have 4a + 2b + c = 14, and with n = 3 we have 9a + 3b + c = 29."
How do u find the formula.. i mean what are the steps u do to find it..can u show how..cuz m not getting it.. :/
Firstly, a is equal to the second difference divided by 2. The first differences are 9, 15, 21, 27, ... so the second difference is 6. This means that a = 6/2 = 3.
So the formula looks like 3n^2+bn+c.
Now you need to find b and c by substituting in some values:
n=1 : The 1st term is 5 so we have 3(1^2) + b(1) + c = 5 --> 3 + b + c = 5 --> b + c = 2
n=2: The 2nd term is 14 so we have 3(2^2) + b(2) + c = 14 --> 12 + 2b + c = 14 --> 2b + c = 2
Now solve these two simultaneous equations (subtract the equations) to find b and c and you get:
b=0, c=2
So the nth term is 3n^2 + 2
An IGCSE question would never ask you to find the nth term of 5, 14, 29, 50, 77, ... without any extra information.