# nth terms!!!

This discussion is closed.
#1
im stuck i cant work out the rule for this nth term pattern ...please help me its for coursewrk
6 24 54 96 150 << there the numbers
0
18 years ago
#2
(Original post by xxclairexx04)
im stuck i cant work out the rule for this nth term pattern ...please help me its for coursewrk
6 24 54 96 150 << there the numbers
is it Un = 6n^2?
0
18 years ago
#3
(Original post by xxclairexx04)
im stuck i cant work out the rule for this nth term pattern ...please help me its for coursewrk
6 24 54 96 150 << there the numbers
\ / \ / \ / \ /
18 30 42 54 First difference
\ / \ / \ /
12 12 12 Second difference

So it's 0.5(12)n^2
= 6n^2

Ye-he-hes!!!
0
#4
(Original post by ZJuwelH)
\ / \ / \ / \ /
18 30 42 54 First difference
\ / \ / \ /
12 12 12 Second difference

So it's 0.5(12)n^2
= 6n^2

Ye-he-hes!!!

what does the ^ stand for
0
18 years ago
#5
(Original post by xxclairexx04)
what does the ^ stand for
'To the power of'. So 6n^2 reads as 6n squared.
0
18 years ago
#6
So what's this courswork then? Is is squares in rectangles? if so look at the KEFW thread that has virtually the full solution and beyond.
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#7
(Original post by Phil_C)
So what's this courswork then? Is is squares in rectangles? if so look at the KEFW thread that has virtually the full solution and beyond.

Ireally dont understand wot nee1 is going on bout
0
18 years ago
#8
...So do I get rep?
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#9
(Original post by ZJuwelH)
...So do I get rep?

whats rep ???? ..and could ne1 ezplain to me as easy as possible lol
0
18 years ago
#10
(Original post by xxclairexx04)
whats rep ???? ..and could ne1 ezplain to me as easy as possible lol
Rep short for reputation, you can give someone positive or negative rep for a post by clicking the yellow box below the post (next to the edit button).

Do you mean you want the answer explained properly?
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#11
(Original post by ZJuwelH)
Rep short for reputation, you can give someone positive or negative rep for a post by clicking the yellow box below the post (next to the edit button).

Do you mean you want the answer explained properly?

Yes please if you dont mind
0
18 years ago
#12
(Original post by xxclairexx04)
Yes please if you dont mind
Basically you find the differences between the terms. If they aren't the same you find the differences of these until they are the same. If you have found the dth difference to always be m then you raise n to the power of d and multiply by m/d, getting m/d(n)^d. Then bung in a value of n and look at the difference between it and the nth term you are looking at. Add or subtract a relevant number accordingly.

In this example we have a second difference of 12. So
m/d(n)^d=12/2(n)^2
=6n^2.
Now if we put in n=1 and compare to the first term, we get 6(1)^2=6. This is the same as the first term (where n=1) and throughout the series so we do not need to add or subtract any number, we have found the formula.

Hope it makes sense, sorry if it don't!
0
18 years ago
#13
(Original post by ZJuwelH)
Basically you find the differences between the terms. If they aren't the same you find the differences of these until they are the same. If you have found the dth difference to always be m then you raise n to the power of d and multiply by m/d, getting m/d(n)^d. Then bung in a value of n and look at the difference between it and the nth term you are looking at. Add or subtract a relevant number accordingly.

In this example we have a second difference of 12. So
m/d(n)^d=12/2(n)^2
=6n^2.
Now if we put in n=1 and compare to the first term, we get 6(1)^2=6. This is the same as the first term (where n=1) and throughout the series so we do not need to add or subtract any number, we have found the formula.

Hope it makes sense, sorry if it don't!
How would you solve it if the number of terms given is not sufficient for a constant common difference to be found?
0
18 years ago
#14
(Original post by Ralfskini)
How would you solve it if the number of terms given is not sufficient for a constant common difference to be found?
Don't think that sort of question would come up...
Haven't got a scooby if that does happen, been a while since I did P1 you see.
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#15
(Original post by ZJuwelH)
Basically you find the differences between the terms. If they aren't the same you find the differences of these until they are the same. If you have found the dth difference to always be m then you raise n to the power of d and multiply by m/d, getting m/d(n)^d. Then bung in a value of n and look at the difference between it and the nth term you are looking at. Add or subtract a relevant number accordingly.

In this example we have a second difference of 12. So
m/d(n)^d=12/2(n)^2
=6n^2.
Now if we put in n=1 and compare to the first term, we get 6(1)^2=6. This is the same as the first term (where n=1) and throughout the series so we do not need to add or subtract any number, we have found the formula.

Hope it makes sense, sorry if it don't!

so do u mean.....6x1=6.......wudnt i have 2 times by 2 ???coz u rit ^2
0
18 years ago
#16
depends on where the sequence of numbers has come from. If it is from a build up of say poctorial patterns, like many GCSE maths coursworks do, then create the next few terms to investigate the differences between them and the differences of the differences etc. From there you can work out the type of power sequence you are dealing with.

Common first difference of "a" then Linear sequence of the form an where "n" is the position
Common 2nd difference of "a" then a quadratic sequence of (a/2)n^2
Common 3rd difference the cubic of form (a/6)n^3
etc

If the sequence is given as a set of numbers and you have found no common difference at the first few levels, 1st, 2nd 3rd etc, you can use a general formulae eg given 5 terms the sequnce could at most be n^4

Let the 5 terms by Z, Y, X, W, and V
then use an^4+bn^3+Cn^2+dn+e=... Z when n=1, Y when n=2 X when n=3 etc. 5 independent equations 5 unknowns so solvable.
0
18 years ago
#17
(Original post by xxclairexx04)
so do u mean.....6x1=6.......wudnt i have 2 times by 2 ???coz u rit ^2
Do you mean if n=2? If so...
6n^2=6(2)^2
=6x4
=24

If that's not what you mean please clarify...
0
18 years ago
#18
(Original post by Phil_C)
depends on where the sequence of numbers has come from. If it is from a build up of say poctorial patterns, like many GCSE maths coursworks do, then create the next few terms to investigate the differences between them and the differences of the differences etc. From there you can work out the type of power sequence you are dealing with.

Common first difference of "a" then Linear sequence of the form an where "n" is the position
Common 2nd difference of "a" then a quadratic sequence of (a/2)n^2
Common 3rd difference the cubic of form (a/6)n^3
etc

If the sequence is given as a set of numbers and you have found no common difference at the first few levels, 1st, 2nd 3rd etc, you can use a general formulae eg given 5 terms the sequnce could at most be n^4

Let the 5 terms by Z, Y, X, W, and V
then use an^4+bn^3+Cn^2+dn+e=... Z when n=1, Y when n=2 X when n=3 etc. 5 independent equations 5 unknowns so solvable.
...What he said.
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#19
(Original post by ZJuwelH)
Do you mean if n=2? If so...
6n^2=6(2)^2
=6x4
=24

If that's not what you mean please clarify...
than kyou sooooooooooooo much i fianly get it , ur better explaing then me teacher , thna alot sorry for taking up ur time
0
18 years ago
#20
(Original post by xxclairexx04)
than kyou sooooooooooooo much i fianly get it , ur better explaing then me teacher , thna alot sorry for taking up ur time
Better than your teacher? Thanks heaps, I was hoping to become a teacher in my middle age...
'Twas my pleasure. Enjoy your work young Claire!
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