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    (Original post by IWasAnEagle)
    anyone know what 'sequences defined iteratively' (on the syllabus) means?

    It means using sigma notation.

    Which I'm just about to ask a question about!

    How did they go from the top line to the second top line?
    I'm quite unfamiliar with sigma notation.. so praying it won't come up
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    (Original post by Noble.)
    You're given a starting value, say a_0, and a function a_{n+1} = f(a_n) for determining further values. A good example of this are Fibonacci numbers, as they're defined to be the sum of the previous two values.
    I'm wrong then, it's not sigma notation. Ooops. I just kind of assumed that.
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    (Original post by jud1th)
    NOOOOOOOOOOO :'(
    Can we just forget it happened instead and wait for rejections/interviews



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    We could just make a mark scheme and you don't have to go on it
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    (Original post by jadoreétudier)
    It means using sigma notation.

    Which I'm just about to ask a question about!

    How did they go from the top line to the second top line?
    I'm quite unfamiliar with sigma notation.. so praying it won't come up
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Views: 245
Size:  7.2 KB
    \displaystyle\sum_{k=0}^m 1 = \underbrace{\textstyle 1 + 1 + ...  +1}_{\mathclap{ \text{ m+1 times }}} = m+1

    \displaystyle\sum_{k=0}^m k = 1 + 2 + 3 + ... + m = \frac{1}{2}m(m+1)
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    (Original post by jadoreétudier)
    It means using sigma notation.

    Which I'm just about to ask a question about!

    How did they go from the top line to the second top line?
    I'm quite unfamiliar with sigma notation.. so praying it won't come up
    Name:  ....png
Views: 245
Size:  7.2 KB
    \displaystyle\sum_{r=1}^n 1 = 1+1+1+. . . +1=n
    and the standard result for:
    \displaystyle\sum_{r=1}^n = \dfrac{n(n+1)}{2} Applying this is how this step is made.

    EDIT: Ninja-ed :ninja:
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    (Original post by gavinlowe)
    .
    (Original post by Oxford Computer Science Dept)
    It's normally the original scripts that are marked. The scripts are marked by real life human beings. Think for a moment about the poor tired marker, wading through lots and lots and lots of papers. Having the answer to a question scattered around an answer paper isn't ideal. They aren't going to have the time to go hunting. But if you you've clearly labelled where you've continued with an answer, they'll look there. They are looking to give you credit for what you've done, not catch you out.

    I know it's easy for me to say, but try to relax. Go and grab yourself a cuppa and then get a good nights sleep. Let your school/test centre worry about the mechanics of it all, and just concentrate, tomorrow, on doing your best. :-)
    One final thing:

    If we find that we need more space than that provided including the deliberately made blank pages, should we use the space given for the 2 questions we aren't meant to attempt and indicate so, or get separate paper from an invigilator?
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    What's everybody doing on their final evenings before the big test? I'm going through the 7 past papers I've made over the past weeks and looking at my mistakes. Stress levels are rising exponentially.
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    (Original post by nahomyemane778)
    A pack of cards consists of 52 dierent cards. A malicious dealer changes one of the cards
    for a second copy of another card in the pack and he then deals the cards to four players,
    giving thirteen to each. The probability that one player has two identical cards is
    (a) 3/13
    (b) 12/51
    (c) 1/4
    (d) 13/51

    I dont understand their solution either- can anyone help?
    It's 12/51. Suppose the first of the duplicated cards is on the top of the pack, and that we give the first 13 cards to the first person, and so on. Now of the 51 cards under the top card, if the duplicate card is within the first 12 cards, it will be given to the same person.
    (I just said the first of the duplicate pair was at the top of the pack for simplicity, but by symmetry it doesn't matter where in the pack it occurs)

    Another explanation is to consider the possible ways in which the 2 cards can occur throughout the pack. There's 52C2 ways of choosing the positions in the pack for the 2 cards to go. Now consider the ways in which the cards can occur within the same person's hand: there's 13C2 ways of choosing two positions in each players hand, and since there's 4 players, that's 4 x 13C2 ways.

    Thus the probability is (4 x 13C2) / (52C2) = 12/51.
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    I told my friend to 'yolo it' and that is precisely what I will do tomorrow.

    #yolo
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    (Original post by yl95)
    I told my friend to 'yolo it' and that is precisely what I will do tomorrow.

    #yolo
    Exactly, so you better get it right this time around.
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    (Original post by jadoreétudier)
    It means using sigma notation.

    Which I'm just about to ask a question about!

    How did they go from the top line to the second top line?
    I'm quite unfamiliar with sigma notation.. so praying it won't come up
    Name:  ....png
Views: 245
Size:  7.2 KB
    I have a slightly different (although equivalent) take.

    We know 4m = 4k + 2j + i.

    If we fix k as 0 first, then j can be anything between 0 and 2m (and the rest can be filled up with i). That's 2m + 1 possibilities.
    If we next fix k as 1, then j can be anything between 0 and 2m - 2. That's 2m - 1 possibilities.
    When k = 2, j has 2m - 3 possibilities.

    We get the arithmetic sequence (2m + 1) + (2m - 1) + (2m - 3) + ... + 1
    There's m + 1 terms in the sequence because k could have taken any value between 0 and m (that's m + 1 values).

    So for our arithmetic series, a = 2m + 1, d = -2 and n = m+1.
    Using the summation formula, we get (m + 1)^2 as desired.

    A similar approach can be taken with SMC/Olympiad problems where we're interested in the number of ways of making up some monetary amount using certain coins.
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    (Original post by gavinlowe)
    The paper gets photocopied, and marked by both universities. (I'm not convinced this is sensible, but it's what is happening...)
    Thank you very much for the answer
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    (Original post by gavinlowe)
    The paper gets photocopied, and marked by both universities. (I'm not convinced this is sensible, but it's what is happening...)
    If as a result of this our papers are going to be photocopied, will pencil still show up?

    Because in the past you have been asked to sketch curves or some form of diagram in a question, and if we did it in pen and wanted to change it (there is no spare copy of a diagram) so you would lose out. If pencil doesn't show up...you can see this would be an issue. So I just wanted to check.

    Thank you
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    (Original post by daniyalfaiz)
    congrats! thats amazing im getting nervous about warwick now :/ what were ur predicted grades/gcse's?
    Thanks!

    Don't worry - I'm sure they just haven't got around to it yet. My predicted grades were 5 A*s (incl. maths and further maths) and I had 10 A*s, 1 A at GCSE, but as long as you're targeted at least AAB and meet the basic uni GCSE requirements, you should get an offer
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    (Original post by IceKidd)
    If as a result of this our papers are going to be photocopied, will pencil still show up?

    Because in the past you have been asked to sketch curves or some form of diagram in a question, and if we did it in pen and wanted to change it (there is no spare copy of a diagram) so you would lose out. If pencil doesn't show up...you can see this would be an issue. So I just wanted to check.

    Thank you
    Will you stop worrying? You are a ridiculously strong candidate - you're not going to get rejected with a score in the high 80s/90s because of presentation, and who ever photocopies them will I am sure ensure they do it well.

    You will be fine
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    I'm turning up with a pen guys - no pencil, rubber or ruler
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    Right now, I wish I'd applied to London Met
    EDIT: Seriously though, I hate this so much
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    I'd be happy with 75 unless we get a nice question 5. Question 5 I'm currently expecting 5/15 on.
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    is it just me, or are these the same as the AEA?
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    (Original post by pirateship)
    is it just me, or are these the same as the AEA?
    Not in my opinion. I think the AEA is just messy - not as many 'tricks' involved!
 
 
 
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