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    Hi does anyone know please whether all the mad as maths questions under ellipse are in the spec? http://madasmaths.com/archive/maths_...c_sections.pdf Thanks
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    (Original post by economicss)
    Hi does anyone know please whether all the mad as maths questions under ellipse are in the spec? http://madasmaths.com/archive/maths_...c_sections.pdf Thanks
    Ellipses? You don't do ellipses in FP1, do you?
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    (Original post by Zacken)
    Ellipses? You don't do ellipses in FP1, do you?
    I don't think so, I think I'm paranoid haha, is it just parabolas and rectangular hyperbola that we need to be able to do for FP1? Thanks
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    (Original post by economicss)
    I don't think so, I think I'm paranoid haha, is it just parabolas and rectangular hyperbola that we need to be able to do for FP1? Thanks
    Yep, just those two are on spec.
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    Thank you!
    (Original post by Zacken)
    Yep, just those two are on spec.
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    http://madasmaths.com/archive/maths_...c_sections.pdf
    http://madasmaths.com/archive/maths_..._induction.pdf
    http://madasmaths.com/archive/maths_...n_practice.pdf
    http://madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_1.pdf
    http://madasmaths.com/archive/maths_..._questions.pdf
    http://madasmaths.com/archive/maths_...ons_part_a.pdf

    Here are all the relevant (to FP1) madasmaths booklets which I was able to pick out! Thought the direct links would be more useful than trying to search for them all manually. Maybe there are a few more I have missed, but these are what I have found! Many thanks to TeeEm - These work booklets are absolutely amazing revision for someone such as myself who has finished all of the actual examination papers! Many thanks!!
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    Hi, I was doing the June 2014 R paper and got stuck on knowing how many assumptions to make for the proof of question 9 (b). Can someone please explain the question to me?
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    (Original post by Glavien)
    Hi, I was doing the June 2014 R paper and got stuck on knowing how many assumptions to make for the proof of question 9 (b). Can someone please explain the question to me?
    You need only assume that:

    u_{k} = 4^{k+1} - 2^{k+3} and u_{k-1} = 4^k - 2^{k+2}.

    Then: u_{k+1} = 6u_{k} - 8u_{k-1} = \cdots where you substitute in from above.

    Since it's a recurrence relation involving three terms, you need to make two assumptions. When there is only two terms, you make 1 assumption, etc...
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    (Original post by Zacken)
    You need only assume that:

    u_{k} = 4^{k+1} - 2^{k+3} and u_{k-1} = 4^k - 2^{k+2}.

    Then: u_{k+1} = 6u_{k} - 8u_{k-1} = \cdots where you substitute in from above.

    Since it's a recurrence relation involving three terms, you need to make two assumptions. When there is only two terms, you make 1 assumption, etc...
    Gotcha, thanks!
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    (Original post by Glavien)
    Gotcha, thanks!
    Cheers.
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    Just under 3 weeks left . . .
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    Hi, please could anyone explain question 4e on exercise A here https://644625398389466aee0063322305...0NRNU0/CH5.pdf I can't really make sense of SB, thanks
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    (Original post by economicss)
    Hi, please could anyone explain question 4e on exercise A here https://644625398389466aee0063322305...0NRNU0/CH5.pdf I can't really make sense of SB, thanks
    Okay, so - first off, you can see in includes just odd numbers - in fact, it's two consecutive odd numbers multiplied together.

    So, the general term is: (2n+1)(2n+3), you want to add up these except starting from 3 * 5, so you need to start from n=1.

    Hence: \sum_{r=1}^{k} (2n+1)(2n+3) = (3)(5) + (5)(7) + \cdots+ (2k+1)(2k+3) and there are precisely k terms in this summation, as required.
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    (Original post by Zacken)
    Okay, so - first off, you can see in includes just odd numbers - in fact, it's two consecutive odd numbers multiplied together.

    So, the general term is: (2n+1)(2n+3), you want to add up these except starting from 3 * 5, so you need to start from n=1.

    Hence: \sum_{r=1}^{k} (2n+1)(2n+3) = (3)(5) + (5)(7) + \cdots+ (2k+1)(2k+3) and there are precisely k terms in this summation, as required.
    Ah I see, brilliant, thank you
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    (Original post by economicss)
    Ah I see, brilliant, thank you
    Cheers.
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    Hey guys,

    I am really stuck on practice paper a Qu 5 in this link:

    http://www.thestudentroom.co.uk/atta...7&d=1244566570

    Surely it is impossible to part a before part b as u can't find the real root
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    (Original post by J.M.Keynes)
    Hey guys,

    I am really stuck on practice paper a Qu 5 in this link:

    http://www.thestudentroom.co.uk/atta...7&d=1244566570

    Surely it is impossible to part a before part b as u can't find the real root
    Huh? You know two roots and you know the constant term.

    You know that if three roots are of a cubic are \alpha,  \beta, \gamma then \alpha \beta \gamma = \frac{10}{2} (in this case).

    You know \alpha \beta = (3+i)(3-i) so \gamma = \frac{10}{2\alpha \beta}
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    How can I expand this out if A and B are both matrices? I know that the order of multiplying matrices matters, so is there a particular way to know which way to order the matrices?
    \mathbf{M} = (\mathbf{A+\mathbf{B}})(2\mathbf  {A-\mathbf{B}})

    Its from the June 2014 R paper, question 6.
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    (Original post by Glavien)
    How can I expand this out if A and B are both matrices? I know that the order of multiplying matrices matters, so is there a particular way to know which way to order the matrices?
    \mathbf{M} = (\mathbf{A+\mathbf{B}})(2\mathbf  {A-\mathbf{B}})

    Its from the June 2014 R paper, question 6.
    You could just work out each bracket numerically and multiply, but if you do expand keep the order the same from left to right, so (A + B)(2A - B) = 2A^2 +2BA - AB - B^2
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    (Original post by NotNotBatman)
    You could just work out each bracket numerically and multiply, but if you do expand keep the order the same from left to right, so (A + B)(2A - B) = 2A^2 +2BA - AB - B^2
    Thanks!


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