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nth terms!!!
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xxclairexx04
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#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
6 24 54 96 150 << there the numbers
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Ralfskini
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#2
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#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
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
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Juwel
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#3
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#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
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!!!
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xxclairexx04
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#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!!!
\ / \ / \ / \ /
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
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Juwel
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#5
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#5
(Original post by xxclairexx04)
what does the ^ stand for
what does the ^ stand for
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Phil_C
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#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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xxclairexx04
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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.
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
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Juwel
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#8
xxclairexx04
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#9
(Original post by ZJuwelH)
...So do I get rep?
...So do I get rep?
whats rep ???? ..and could ne1 ezplain to me as easy as possible lol
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Juwel
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#10
(Original post by xxclairexx04)
whats rep ???? ..and could ne1 ezplain to me as easy as possible lol
whats rep ???? ..and could ne1 ezplain to me as easy as possible lol
Do you mean you want the answer explained properly?
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xxclairexx04
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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?
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
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Juwel
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#12
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#12
(Original post by xxclairexx04)
Yes please if you dont mind
Yes please if you dont mind
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!
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Ralfskini
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#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!
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!
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Juwel
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#14
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#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?
How would you solve it if the number of terms given is not sufficient for a constant common difference to be found?
Haven't got a scooby if that does happen, been a while since I did P1 you see.
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xxclairexx04
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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!
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
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Phil_C
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#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.
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.
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Juwel
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#17
(Original post by xxclairexx04)
so do u mean.....6x1=6.......wudnt i have 2 times by 2 ???coz u rit ^2
so do u mean.....6x1=6.......wudnt i have 2 times by 2 ???coz u rit ^2
6n^2=6(2)^2
=6x4
=24
If that's not what you mean please clarify...
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Juwel
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#18
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#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.
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.
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xxclairexx04
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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...
Do you mean if n=2? If so...
6n^2=6(2)^2
=6x4
=24
If that's not what you mean please clarify...
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Juwel
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#20
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#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
than kyou sooooooooooooo much i fianly get it , ur better explaing then me teacher , thna alot sorry for taking up ur time
'Twas my pleasure. Enjoy your work young Claire!
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