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    (Original post by chaosdestro0)
    Very small chance you will need to know the proof of sum of series.
    I am just trying to find it right now.
    Edit:
    Arithmetic proof

    Sn = a + (a + d) + (a + 2d) + (a + 3d) + ..... (a + (n-2)d) + (a + (n-1)d)
    Reversed Sn = (a + (n-1)d) + (a + (n-2)d) .... (a+ 3d) + (a+2d) + (a + d) + a

    Adding Sn to reversed Sn

    2Sn = 2a + (n - 1)d + 2a + (n-1)d + 2a(n-1)d

    2sn = n[2a + (n-1)d]
    Sn = n/2[2a + (n-1)d]
    You clearly haven't looked at many papers if you think that. It has come up in the past and could easily come up again. I'm kinda hoping stuff like that does come up to catch some people out, although... I'm going for full marks so I don't really think people getting stuff right should affect me. Beginning to think C1 and C2 might be a bit of fun, I'll see if I still think that after I've completed the three hours.
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    Best Website To Revise Off
    Is Exam Solutions!
    Its Like GCSE Maths Watch
    But Its Quite Slow Now
    Bcz Too Many Ppl Are Revisin
    Off It Nw!
    But In All...Great Resource!!
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    Edexcel scum with exam solutions :P
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    (Original post by Edwin Okli)
    You clearly haven't looked at many papers if you think that. It has come up in the past and could easily come up again. I'm kinda hoping stuff like that does come up to catch some people out, although... I'm going for full marks so I don't really think people getting stuff right should affect me. Beginning to think C1 and C2 might be a bit of fun, I'll see if I still think that after I've completed the three hours.
    I have done many past papers and have yet to see this, can you give me an example of this question.
    have checked : 2009 c1 June and january nope
    2010 june- Nope
    2006 January- Nope
    2007 June-Nope

    I have done a fairly large amount of papers and have not seen a questions asking for proof as far as I can remember
    Just remember that I started revision in november so it is fairly long ago.
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    (Original post by chaosdestro0)
    I have done many past papers and have yet to see this, can you give me an example of this question.
    June 2005
    Q9a) Prove that the sum of the first n terms of the series is 1/2 n[2a + (n ?1)d].
    First example I found, not sure if it did come up again after that. But that doesn't mean it isn't likely to come up tomorrow.
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    Good luck for the exam tomorrow people. Hope all goes well.
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    Good luck everyone!!
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    @people who helped with formula's

    Thanks Guys

    ____________

    Like most of you I am really worried about making silly mistakes.

    I make the most stupid mistakes and lost 15 marks in the specimen because of that, but then in some papers I make hardly any silly mistakes!
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    Good luck everyoneeee!
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    BTW does anyone have links to how your prove those arithmetic/geometric series formula's?!!!
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    (Original post by Darthdevidem)
    BTW does anyone have links to how your prove those arithmetic/geometric series formula's?!!!
    If you've got an arithmetic series with n terms

    a, (a+d), (a+2d), (a+3d), ..., [a+(n-2)d], [a+(n-1)d]

    Call the sum of these terms S_n

    So

    S_n = a + (a+d) + (a+2d) + (a+3d) + ... + [a+(n-2)d] + [a+(n-1)d] \ \ \ \ \ \ast_1

    If you rewrite the same sum, but in reverse, you get

    S_n = [a+(n-1)d] + [a+(n-2)d] + ... + (a+3d) + (a+2d) + (a+d) + a \ \ \ \ \ \ast_2

    If you do \ast_1 + \ast_2 you can see the terms are identical.

    By this I mean, you end up with:

    a + a+(n-1)d = 2a + (n-1)d
    a+d + a + (n-2)d = 2a + (n-1)d

    The terms are symmetrical, you're increasing the common difference with \ast_1 and you're decreasing the common difference with \ast_2. Because there are n terms, you end up with n many 2a+(n-1)d

    So

    2S_n = n[2a + (n-1)d]
    S_n = \frac{n}{2}[2a + (n-1)d]





    For a geometric series, if you've got the series with n terms

    a, ar, ar^2, ar^3, ..., ar^{n-2}, ar^{n-1}

    Then the sum of these n terms, S_n

    S_n = a + ar + ar^2 + ar^3 + ... + ar^{n-2} + ar^{n-1} \ \ \ \ \ \ast_3

    If you multiply \ast_3 by r

    rS_n = ar + ar^2 + ar^3 + ar^4 + ... + ar^{n-1} + ar^n

    You can see that all the terms of rS_n are identical to those of S_n bar the a in S_n and the ar^n in rS_n

    So if you subtract the second equation from \ast_3 you end up with

    S_n(1-r) = a - ar^n

    S_n = \frac{a-ar^n}{1-r}


    I might as well go on to explain how you obtain the sum to infinity formula from this, for when -1 < r < 1

    If you got the formula

    S_n = \frac{a-ar^n}{1-r}

    and you're summing it to infinity:

    n = \infty

    S_\infty = \frac{a-ar^\infty}{1-r}

    Providing -1 < r < 1, r^\infty = 0, you can try this on your calculator, the value of r will tend to zero as you keep on multiplying it by r.

    So, because r^\infty = 0, ar^\infty = 0 and so

    S_\infty = \frac{a}{1-r}
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    {mod: exam discussion removed}
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    You cannot discuss this exam until midnight. Please read the related announcement.
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    Mod: exam discussion removed.
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    My bad, I thought we could discuss the hardness of it lol x
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    The announcement is very clear:

    This means no discussion AT ALL - not even "I found it hard/easy" or "I dropped my pencil". The line between

    * "I found it hard" and
    * "I found question 2 hard" and
    * "I found question 2 about <insert subject> hard"
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    You cannot discuss this exam until midnight.
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    mod: text removed
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    cant discuss it .
 
 
 
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