how do you work out the nth term for a sequence where the diference in numbers is different? for example
1-st row: 1, 4, 10, 20, 35, 56, 84, ...
2-nd row: 3, 6, 10, 15, 21, 28, ...
3-rd row: 3, 4, 5, 6, 7, ...
4-th row: 1, 1, 1, 1, ...
after this what do you do next to find the nth term?
Turn on thread page Beta
Nth Term watch
- Thread Starter
- 14-06-2005 22:03
- 14-06-2005 22:11
changing n'th term is no longer on the gcse sylabus took off from 2002?
- 14-06-2005 22:11
Umm, I THINK...that if the first difference is different and you've to move to a second difference it's n^2...
to find the nth term then I *think* you write the square numbers (1, 4, 9, 16) above the sequence numbers and whatever you do to get from the square numbers to the number in the sequence is the second part ie n2 - 3 if you had to subtract 3 to get the number....but to be honest I always get muddled up so this could be wrong...please someone correct me
- 14-06-2005 22:37
use the formula: a+(n-1)+1/2(n-1)(n-2)c
- 14-06-2005 22:49
I do know the method, but it is awfully complex if you want the answer straight away - Here goes nothing...
Ok - you have:
Term (n): 1 2 3 4 5 6 7....
Sequence: 1, 4, 10, 20, 35, 56, 84, ...
1st row: 3, 6, 10, 15, 21, 28, ...
2nd row: 3, 4, 5, 6, 7, ...
3rd row: 1, 1, 1, 1, ...
Now, because we get constant differences in row 3, it will be a cubic equation, and so we will get something in the form an³ + bn² + cn + d.
As a result, we can write everything algebraically.
Where n = 1, the equation for the corresponding term will be:
a + b + c + d
Where n = 2, the equation for the corresponding term will be:
8a + 4b + 2c + d
Where n = 3, the equation for the corresponding term will be:
27a + 9b + 3c + d
Where n = 4, the equation for the corresponding term will be:
64a + 16b + 4c + d (and I'm sure you can see the pattern).
If we take the differences of these, you'll get the algebraic equivalent of the 1st row sequence (as above).
As a result, you'll get the seqeunce:
(7a + 3b + c), (19a + 5b + c), (37a + 7b + c) .... etc.
Now, if we take the differences of these (as you did with the numbers), we get the algebraic equivalent of the 2nd row sequence (as above).
As a result, we get:
(12a + 2b), (18a + 2b)
Now.... If we take the differences of these, we'll finally get the 3rd row sequence, where all values are the same!
So, if we take the difference of these, we get....
6a.... (and it would continue to do so...)
If 6a = 1, therefore a= 1/6.
Now if we continue to substitute (going up from the bottom), we find that b= 1/2
We'll find that c = 1/3, and that d = 0 ( I think...)
Lets have a check...
(8*1/6) + (4*0.5) + (2* 1/3) = 4
Which it does - HURRAH!
As a result, using the first formula i gave you, you can subsititute in the nth term (the one you want to find) with a being 1/6, b being 0.5 and c being 1/3 (i.e. 1/6n³ + 0.5n² + 1/3n = nth term)
Phew - that took a long time (any rep for my hard work?? !)
- 15-06-2005 01:06
Formula for nth Term, whereby the differences change:
a+(n-1)d + 1/2(n-1)(n-2)C
a = The first term
d = The difference between the first two terms (ie. the first difference)
C = the change in the differences (ie. plus two, or minus 3).