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How do u find the nth term of the formulaaa??? HELP!!

Find a formula, in terms of n, for the nth term of the sequence:

1, –3, 5, –7, 9,.....

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Reply 1
Look up how to form quadratic sequences
(edited 5 years ago)
Original post by apaul322
Find a formula, in terms of n, for the nth term of the sequence:

1, –3, 5, –7, 9,.....


Compare with the sequence 1, 3, 5, 7, 9
The nth term of this one is 2n-1. So this needs to be multiplied by (-1) every term.
So let's try (-1)^n * (2n-1)
When n=1, (-1)^1*(2(1)-1) = -1
So we got -1 when we want 1. So let's try (-1)^(n-1) * (2n-1). This seems to work since it's the odd numbers which switch between positive and negative

So the nth term is (-1)^(n-1) * (2n-1)
Original post by BDunlop
Look up how to form quadratic sequences


it's not quadratic

:beard:
Original post by BDunlop
Look up how to form quadratic sequences


This one isn't quadratic since the second difference isn't the same. Since it constantly switches between positive and negative, it's an alternating sequence.
Reply 5
Original post by apaul322
Find a formula, in terms of n, for the nth term of the sequence:

1, –3, 5, –7, 9,.....


So basically use the formula: a+d(n-1)
A= the first term
D=the difference
N=the sequence number.
So for example: 2,5,8,11...(+3)
#to calculate the 11th term:
2+3(11-1)
=2+30
=32
#to calculate the nth term
2+3(n-1)
2+3n-3
=3n-1
BUT this formula is for the basic sepuences
Original post by Ash.Nasa
So basically use the formula: a+d(n-1)
A= the first term
D=the difference
N=the sequence number.
So for example: 2,5,8,11...(+3)
#to calculate the 11th term:
2+3(11-1)
=2+30
=32
#to calculate the nth term
2+3(n-1)
2+3n-3
=3n-1
BUT this formula is for the basic sepuences


Yeah this isn't an arithmetic sequence, there is no common difference, so unfortunately that formula won't work here.
Reply 7
Original post by 3pointonefour
Compare with the sequence 1, 3, 5, 7, 9
The nth term of this one is 2n-1. So this needs to be multiplied by (-1) every term.
So let's try (-1)^n * (2n-1)
When n=1, (-1)^1*(2(1)-1) = -1
So we got -1 when we want 1. So let's try (-1)^(n-1) * (2n-1). This seems to work since it's the odd numbers which switch between positive and negative

So the nth term is (-1)^(n-1) * (2n-1)


got it now thankss:smile:
Original post by apaul322
got it now thankss:smile:


No problem :smile:
Reply 9
Original post by apaul322
Find a formula, in terms of n, for the nth term of the sequence:

1, –3, 5, –7, 9,.....

By inspection we have an alternating sign which is encoded via a factor (1)n (-1)^n , then the sequence is just increasing by 2 each time so: (1)n(2n)(-1)^n \cdot (2n) , but then the sequence beginning at n = 0 is 1 so:

an=(1)n(2n+1)[br]n=0,1, a_n = (-1)^n \cdot (2n + 1)[br]n = 0, 1, \dots
Original post by Ryanzmw
By inspection we have an alternating sign which is encoded via a factor (1)n (-1)^n , then the sequence is just increasing by 2 each time so: (1)n(2n)(-1)^n \cdot (2n) , but then the sequence beginning at n = 0 is 1 so:

an=(1)n(2n+1)[br]n=0,1, a_n = (-1)^n \cdot (2n + 1)[br]n = 0, 1, \dots


Doesn't an nth term, by definition, start at n = 1?
Original post by 3pointonefour
Doesn't an nth term, by definition, start at n = 1?


Not always, in fact in my experience series tend to begin at 0 more often than at 1. Though should you be required to begin at 1 then it's just a simple matter of rewriting (n-1) where the n's are.
Original post by Ryanzmw
Not always, in fact in my experience series tend to begin at 0 more often than at 1. Though should you be required to begin at 1 then it's just a simple matter of rewriting (n-1) where the n's are.


I see. Can you give me examples from experience when series start at 0? I don't doubt you, I'm just genuinely interested to learn.
Original post by 3pointonefour
I see. Can you give me examples from experience when series start at 0? I don't doubt you, I'm just genuinely interested to learn.


Aaah well just in my experience of analysis and in numerical analysis formulae we tend to begin sequences with n = 0. One especially good reason for getting into the habit of thinking like this is if you do any computer programming (which is bound to be a part of any maths/physics/engineering degree) you'll find that computers keep lists with indices beginning at 0. So working with the notion that sequences begin with n = 0 tends to make everything simpler in my opinion.
Original post by Ryanzmw
Aaah well just in my experience of analysis and in numerical analysis formulae we tend to begin sequences with n = 0. One especially good reason for getting into the habit of thinking like this is if you do any computer programming (which is bound to be a part of any maths/physics/engineering degree) you'll find that computers keep lists with indices beginning at 0. So working with the notion that sequences begin with n = 0 tends to make everything simpler in my opinion.


Oh wow! It makes sense since many programming languages use zero-based forms of counting. Thanks! Are you studying at
undergraduate level then?
Original post by 3pointonefour
Oh wow! It makes sense since many programming languages use zero-based forms of counting. Thanks! Are you studying at
undergraduate level then?


I mean it really doesn't matter than much, it's like some people consider the natural numbers to include 0, others don't.

Yep message me if you want to ask anything about the material!
Original post by Ryanzmw
I mean it really doesn't matter than much, it's like some people consider the natural numbers to include 0, others don't.

Yep message me if you want to ask anything about the material!


What degree are you studying for? And how big is the jump from A-level to uni in terms of content? Is there more emphasis on different stuff etc.
Original post by 3pointonefour
What degree are you studying for? And how big is the jump from A-level to uni in terms of content? Is there more emphasis on different stuff etc.


I'm studying for a Mmath (i.e. 4 year maths). A-level to first year isn't a huge jump provided you do further maths and remember the content. The only parts of first year that people struggle with is real analysis (which is not bad once you get used to doing pure maths) and even showing up to the lectures.
Original post by Ryanzmw
I'm studying for a Mmath (i.e. 4 year maths). A-level to first year isn't a huge jump provided you do further maths and remember the content. The only parts of first year that people struggle with is real analysis (which is not bad once you get used to doing pure maths) and even showing up to the lectures.


Anything you recommend for reading/learning? I'm in year 11 but ive self taught A-level maths and FM. Not sure where to go from there. Any advice on what to do to learn more maths?
Original post by 3pointonefour
Anything you recommend for reading/learning? I'm in year 11 but ive self taught A-level maths and FM. Not sure where to go from there. Any advice on what to do to learn more maths?


A good book is https://www.amazon.co.uk/Bridging-University-Mathematics-Edward-Hurst/dp/1848002890

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