reb0xx
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on page 94 it says
(You could be asked to prove these formula) What does it mean

its the formula for the sum of an arithmetic series
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thecookiem0nster
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It means in the C1 Exam a question could say derive the formula for the sum of a series. You must be able to start from simple equations and work your way to finding the equation for the sum. You must be able to show it all on paper too
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nuodai
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It means that you could be asked to show, for example, that the nth term of an arithmetic sequence with initial term a and common difference d is a+(n-1)d, or that the nth partial sum is na + \dfrac{n(n-1)}{2}d, etc... that is, you might have to derive these results rather than quoting them.
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ilyking
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just memorise the way they derive. It's quite easy actually, consider it a gift if it does come up
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im so academic
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(Original post by ilyking)
just memorise the way they derive. It's quite easy actually, consider it a gift if it does come up
What if they word it differently?
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Instincts_2012
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(Original post by im so academic)
What if they word it differently?
Then it depends on how smart OP or other people are to spot what the question means + think outside the box and find other methods of deriving it
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ilyking
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(Original post by im so academic)
What if they word it differently?
They can't really, if they ask you to derive you just derive. You can't really twist it and make it complicated if you know what I mean

if it does ask you to derive but not directly telling you, just spot it and then derive
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tiny hobbit
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(Original post by nuodai)
It means that you could be asked to show, for example, that the nth term of an arithmetic sequence with initial term a and common difference d is a+(n-1)d, or that the nth partial sum is na + \dfrac{n(n-1)}{2}d, etc... that is, you might have to derive these results rather than quoting them.
The version of the sum of n terms most often used in C1 is 0.5n(2a+(n-1)d)
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im so academic
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(Original post by tiny hobbit)
The version of the sum of n terms most often used in C1 is 0.5n(2a+(n-1)d)
That's what I thought. I don't think the other formula is the "Edexcel" version.
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Nkhan
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What exam board is this?
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nuodai
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(Original post by tiny hobbit)
The version of the sum of n terms most often used in C1 is 0.5n(2a+(n-1)d)
They're the same, so I doubt it matters :p: I find writing it as na + \dfrac{n(n-1)}{2}d easier to remember (and easier to derive), since it emphasises that you've got n lots of the initial term and (1+2+...+n-1) lots of the common difference. The factorised form sort of garbles this information.
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tiny hobbit
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(Original post by nuodai)
They're the same, so I doubt it matters :p: I find writing it as na + \dfrac{n(n-1)}{2}d easier to remember (and easier to derive), since it emphasises that you've got n lots of the initial term and (1+2+...+n-1) lots of the common difference. The factorised form sort of garbles this information.
Yes they're the same, but I thought that it would be helpful in a thread about C1 to give the version that C1 students would be familiar with and that they get given in their formula book.
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perrytheplatypus
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So...how exactly do you derive it from 0.5n(a+l)?
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tiny hobbit
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(Original post by perrytheplatypus)
So...how exactly do you derive it from 0.5n(a+l)?
l is the last term i.e. the nth term which is a+(n-1)d. Replace l with this and you're there. I take it you have derived 0.5(a+l)?
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perrytheplatypus
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(Original post by tiny hobbit)
l is the last term i.e. the nth term which is a+(n-1)d. Replace l with this and you're there. I take it you have derived 0.5(a+l)?
No, that's the bit I can't do
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tiny hobbit
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(Original post by perrytheplatypus)
No, that's the bit I can't do
Sn = a + (a+d) + (a+2d).......+(a+(n-1)d)
Then reverse it:
Sn = (a+(n-1)d) +..................+a
When you add the 2 lines together you get
2Sn = (2a+(n-1)d)+(2a+(n-1)d)..... (n times)
=n(2a+(n-1)d)
Then divide by 2
Sn = 0.5n(.....)
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perrytheplatypus
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Ah, I see...though I wish we did sequences in class...>_<
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tiny hobbit
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(Original post by perrytheplatypus)
Ah, I see...though I wish we did sequences in class...>_<
Which board are you doing? Are you sure that Arithmetic Series are in your specification? They are in Edexcel C1 but not until C2 for OCR (I haven't checked the others).
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MostCompetitive
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They'll say something like this: prove that the sum of the first n terms of an arithmetic series with first term a and common difference d is given by \frac12n[2a+(n-1)d] (or \frac{n}{2}[2a+(n-1)d]). It is a simple proof.
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perrytheplatypus
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(Original post by tiny hobbit)
Which board are you doing? Are you sure that Arithmetic Series are in your specification? They are in Edexcel C1 but not until C2 for OCR (I haven't checked the others).
I'm doing Edexcel...
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