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    (Original post by Reeyap)
    How would you do part d?
    You would find each of the gradients by using the formula y(2)-y(1)/x(2)-x(1) and then form an equation in which the product of this gradients equals -1 and then solve for p^2q^2
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    Hi, I am stuck on question 10(ii) of the June 2013 R paper. I understand that summation of the r^2 term and the -2r term, but don't understand what the summation of \sum\limits_{r=0}^n (2n+1) will produce. I initially thought it will be equal to n(2n+1) but I am pretty sure that's incorrect. Does the summation from 0 matter or influence the result?
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    (Original post by Patrick2810)
    \sum\limits_{r=0}^n (2n+1) = \sum\limits_{r=1}^n (2n+1) + 2(0) +1

    I believe
    How did you get to that answer?
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    (Original post by Glavien)
    How did you get to that answer?
    Sorry my mistake - not sure what do you since you aren't summing "r" terms

    do you think it might be n(2n+1) + 1 ?
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    (Original post by Patrick2810)
    Sorry my mistake - not sure what do you since you aren't summing "r" terms

    do you think it might be n(2n+1) + 1 ?
    I thought it would be n(2n+1).
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    (Original post by Glavien)
    I thought it would be n(2n+1).
    when n=0, 2n+1 = 1 though?
    then from n=1 to n=n it's a sum of n times 2n+1

    hence n(2n+1) + 1

    Not sure though
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    (Original post by Glavien)
    Hi, I am stuck on question 10(ii) of the June 2013 R paper. I understand that summation of the r^2 term and the -2r term, but don't understand what the summation of \sum\limits_{r=0}^n (2n+1) will produce. I initially thought it will be equal to n(2n+1) but I am pretty sure that's incorrect. Does the summation from 0 matter or influence the result?
    I'll give you two ways to look at it:

    1. \displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

    2. \displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

    Since in this case you have \sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

    Then you have f(r) = 1 being added together n+1 times to get a total of (2n+1)(1 + 1 + \cdots  +1) = (2n+1)(n+1).
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    (Original post by Patrick2810)
    when n=0, 2n+1 = 1 though?
    then from n=1 to n=n it's a sum of n times 2n+1

    hence n(2n+1) + 1

    Not sure though
    Mark Scheme got 2n(n+1) + (n+1)
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    (Original post by Patrick2810)
    when n=0, 2n+1 = 1 though?
    then from n=1 to n=n it's a sum of n times 2n+1

    hence n(2n+1) + 1

    Not sure though
    See my post above. Remember that you are summing over r.

    So when r=0 we have (2n+1) = (2n+1) in no way are we ever setting n=0.
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    (Original post by Zacken)
    See my post above. Remember that you are summing over r.

    So when r=0 we have (2n+1) = (2n+1) in no way are we ever setting n=0.
    ah ok,it's just strange that you're summing for r, but the sum is of "n"s... I guess you just remember to do n * the thing in the sum, and add 1 * the thing in the sum (for when r=0) ?
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    (Original post by Zacken)
    I'll give you two ways to look at it:

    1. \displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

    2. \displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

    Since in this case you have \sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

    Then you have f(r) = 1 being added together n+1 times to get a total of (2n+1)(1 + 1 + \cdots  +1) = (2n+1)(n+1).
    Thanks so much, this really helped! I understand the question using the second way, but the first way is a bit confusing. For the 0th term could you write it as \displaystyle \sum_{r=0}^0 (2n+1) = (2n+1) \sum_{r=0}^0 1 and then isn't \displaystyle \sum_{r=0}^0 1 = 0 , so the whole term is 0?
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    (Original post by Patrick2810)
    ah ok,it's just strange that you're summing for r, but the sum is of "n"s... I guess you just remember to do n * the thing in the sum, and add 1 * the thing in the sum (for when r=0) ?
    Read my answer again, there's no need to remember anything, just sit down and try and understand what's going on with there being (n+1) terms in total and that n is a fixed, unchanging constant integer. Once it clicks, it'll all be clear.
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    (Original post by Glavien)
    Thanks so much, this really helped! I understand the question using the second way, but the first way is a bit confusing. For the 0th term could you write it as \displaystyle \sum_{r=0}^0 (2n+1) = (2n+1) \sum_{r=0}^0 1 and then isn't \displaystyle \sum_{r=0}^0 1 = 0 , so the whole term is 0?
    No, that doesn't make sense, the first term has nothing to do with a sum.

    \displaystyle \sum_{r=0}^n f(r) = f(0) + f(1) + \cdots

    Here, we have f(0) = 2n+1, that's the 0th term. There's no sum coefficient or anything.
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    Has anyone completed the crashmaths FP1 past papers? I feel like there are multiple mistakes in the first one.
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    The only bit annoying me about FP1 are the bloody divisibility proof by induction questions.

    I just hope this exam isn't horrible, but seeing how 2015 didn't have a divisibility proof question...
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    (Original post by Zacken)
    No, that doesn't make sense, the first term has nothing to do with a sum.

    \displaystyle \sum_{r=0}^n f(r) = f(0) + f(1) + \cdots

    Here, we have f(0) = 2n+1, that's the 0th term. There's no sum coefficient or anything.
    Yes, took a while but I think i got it, thank you!!!
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    (Original post by alfmeister)
    Has anyone completed the crashmaths FP1 past papers? I feel like there are multiple mistakes in the first one.
    where do you think the mistakes are?
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    (Original post by alfmeister)
    Has anyone completed the crashmaths FP1 past papers? I feel like there are multiple mistakes in the first one.
    Having completed the paper I, myself, noticed one mistake in the mark scheme of the 'Set A' paper. I did not notice any other mistakes in it apart from one. I think the paper was very good - you can't expect everything to be flawless ultimately, CrashMaths/Aaran has other things to worry about than one FP1 paper, all he's trying to do is make a paper which is giving us an extra opportunity to see where we are at with our knowledge and understanding of the FP1 module. I'd say you should be slightly more grateful.

    If you don't mind me asking, what mistakes did you notice? I only noticed one arithmetic error in the mark scheme which I pointed out and was amended quickly, otherwise apart from that I did not find any others, and found the paper actually very useful..
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    (Original post by iMacJack)
    Having completed the paper I, myself, noticed one mistake in the mark scheme of the 'Set A' paper. I did not notice any other mistakes in it apart from one. I think the paper was very good - you can't expect everything to be flawless ultimately, CrashMaths/Aaran has other things to worry about than one FP1 paper, all he's trying to do is make a paper which is giving us an extra opportunity to see where we are at with our knowledge and understanding of the FP1 module. I'd say you should be slightly more grateful.

    If you don't mind me asking, what mistakes did you notice? I only noticed one arithmetic error in the mark scheme which I pointed out and was amended quickly, otherwise apart from that I did not find any others, and found the paper actually very useful..
    I am by no means ungrateful, I was simply saying that I think I may have found one or two mistakes> However I might be wrong, in the mark scheme of paper one question 3. The algebraic division doesn't seem to be correct.
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    (Original post by kingaaran)
    where do you think the mistakes are?
    In the mark scheme for paper 1, question 3.More specifically the algebraic division.
 
 
 
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