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    (Original post by reubenkinara)
    I wasn't aware there was a math's course work component for any exam board let alone one complimentary for C3!
    OCR MEI Maths C3 has an investigation into iteration as coursework.
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    (Original post by Únie)
    OCR MEI Maths C3 has an investigation into iteration as coursework.
    Interesting. Thanks
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    (Original post by joostan)
    Well, what's d, and a? Subbing those values into the derivation from the start (or the end) can yield the answer - alternatively I could prbably dream up a proof by induction
    No, it's fine...

    a = 1
    d = 4

    n/2 (1 + 4(n-1))

    (n + 4n^2 - 4n)/2
    = 4n^2 - 3n/2

    So I'm confused
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    (Original post by L'Evil Fish)
    No, it's fine...

    a = 1
    d = 4

    n/2 (1 + 4(n-1))


    (n + 4n^2 - 4n)/2
    = 4n^2 - 3n/2

    So I'm confused
    Check the bold - you'll kick yourself when you spot it
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    (Original post by joostan)
    Check the bold - you'll kick yourself when you spot it
    (n-1) should be substituted with the other value they have? (4n-7)??
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    (Original post by L'Evil Fish)
    (n-1) should be substituted with the other value they have? (4n-7)??
    You're over complicating this :lol:
    Spoiler:
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    Check the formula again and look at it again
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    (Original post by joostan)
    You're over complicating this :lol:
    Spoiler:
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    Check the formula again and look at it again
    Sn = n/2 (2a + (n-1)d)
    = n/2 ((2)(1) + (n-1)4)

    Don't get what I've done wrong
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    (Original post by L'Evil Fish)
    Sn = n/2 (2a + (n-1)d)
    = n/2 ((2)(1) + (n-1)4)

    Don't get what I've done wrong
    Precisely.
    What's  2 \times 1? and check your working
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    (Original post by joostan)
    Precisely.
    What's  2 \times 1? and check your working
    :rofl: :rofl: :zomg:

    That'll happen in the exam too...

    n/2 (2 + 4n-4)
    n/2 (-2+4n)
    (-2n+4n^2)/2
    2n^2-2n
    2n(n-1) or even n(2n-1) as they want
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    (Original post by L'Evil Fish)
    ...
    They're not equivalent, I suspect you've made an error somewhere.

    Don't really understand what the question means and what constitutes as deriving from first principles, but alternatively you could always split the sum up to:

    \displaystyle\sum_{r=1}^{n} 4r - 3 = \displaystyle\sum_{r=1}^{n} 4r - \displaystyle\sum_{r=1}^{n} 3 and substitute 4r = \frac{1}{2} ([2r+1]^{2} - [2r-1]^{2} ) which all telescopes to

    \frac{1}{2}((2n+1)^{2} - 1) - 3n and simplify
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    (Original post by L'Evil Fish)
    :rofl: :rofl: :zomg:

    That'll happen in the exam too...

    n/2 (2 + 4n-4)
    n/2 (-2+4n)
    (-2n+4n^2)/2
    2n^2-2n

    2n(n-1) or even n(2n-1) as they want
    I'm sorry to say that the two forms you've given me are not equivalent. Check the bold.
    The inductive proof was so much tidier
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    (Original post by Boy_wonder_95)
    Ouch how many marks did that cost you? I dropped 1 or 2 marks at the end as I forgot to times the probability by 2 :blushing:, but there was a lot of tricky parts!

    This question buggered me:

    Attachment 216178

    I got the right answers using trial and error but what would be a more neater way of solving it?
    As said, it cost me 5 UMS - I don't know exactly how many raw marks that is though, and I'm fairly sure I got the rest of the paper right.

    (Original post by L'Evil Fish)
    I.don't normally do this. But help please
    You're not allowed to just quote the summation formula, so there are a few things you could do. Firstly, you could prove the summation formula  S_n = \frac{1}{2}n(2a+(n-1)d) (I think that's the formula) but in my opinion a better way would be to simply prove the given summation using induction.

    (You could also prove the result for the sum from one to n of r, and then use this and series methods to derive the result. The problem is that I'm not familiar with the WJEC syllabus so I don't know what method is going to be expected.)
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    (Original post by Fuzzy12345)
    Don't really understand what the question means and what constitutes as deriving from first principles, but you could always split the sum up to:

    \displaystyle\sum_{r=1}^{n} 4r - 3 = \displaystyle\sum_{r=1}^{n} 4r - \displaystyle\sum_{r=1}^{n} 3 and substitute 4r = \frac{1}{2} ([2r+1]^{2} - [2r-1]^{2} ) which all telescopes to

    \frac{1}{2}((2n-1)^{2} - 1) - 3n and simplify
    Is this the same problem L'Evil Fish is working on? :eek:
    Check the last few posts.
    There's a proof by induction, but I think that this is satisfactory. Also I'm not convinced that what you've done is from first principles
    EDIT: I realise now that you were trying to help L'Evil Fish
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    (Original post by DJMayes)
    You're not allowed to just quote the summation formula, so there are a few things you could do. Firstly, you could prove the summation formula  S_n = \frac{1}{2}n(2a+(n-1)d) (I think that's the formula) but in my opinion a better way would be to simply prove the given summation using induction.
    (You could also prove the result for the sum from one to n of r, and then use this and series methods to derive the result. The problem is that I'm not familiar with the WJEC syllabus so I don't know what method is going to be expected.)
    The formula is right - induction so much tidier but I'm pretty sure it's on the FP1 spec
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    (Original post by Boy_wonder_95)
    Just did the S1 paper you sat, it was quite greasy...
    Which paper was that? :lol:
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    (Original post by joostan)
    Is this the same problem L'Evil Fish is working on? :eek:
    Check the last few posts.
    There's a proof by induction, but I think that this is satisfactory. Also I'm not convinced that what you've done is from first principles
    EDIT: I realise now that you were trying to help L'Evil Fish
    Yeah...I was trying to help the OP.
    Also really? You've assumed the summation formula in your solution and the question explicitly asks for a solution from first principles and I'd argue that the method of differences is as elementary as you can get - it's just toying around with algebra...
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    (Original post by justinawe)
    Which paper was that? :lol:
    Jan 2012, if he's correct that it's the one I sat.
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    (Original post by Fuzzy12345)
    Yeah...I was trying to help the OP.
    Also really? You've assumed the summation formula in your solution and the question explicitly asks for a solution from first principles and I'd argue that the method of differences is as elementary as you can get - it's just toying around with algebra...
    On second thoughts I suppose you're right.
    Btw I proved the summation formula on the previous page
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    (Original post by DJMayes)
    M2 is easily Jan 2010. I'm not 100% sure with C3 and C2, they all felt pretty much the same to me.
    k then. I'll take that as my one and only mock for M2
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    (Original post by L'Evil Fish)
    Okay C2 question firstly...

    nth term is 4n - 3

    Prove from first principles, without the summation formula that

    Sn = 1 + 5 + ... + (4n-7) + (4n-3)

    = n(2n-1)

    ???
    may be you should just write Sn backwards, add the two, half it, and simplify.
 
 
 
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