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    swear the F1 papers are harder than the IAL FP1 papers??? had a 4+ mark difference between the 2 papers for jan 14
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    (Original post by tazza ma razza)
    swear the F1 papers are harder than the IAL FP1 papers??? had a 4+ mark difference between the 2 papers for jan 14
    Careful now, some of those papers have content that isn't in your FP1 spec (just one teeny bit of content, but that might be what caused the mark drop). However, F1/FP1 papers are not inherently harder than one another, one may be harder iin a particular year as a coincidence, but it is not intended.
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    (Original post by Zacken)
    Careful now, some of those papers have content that isn't in your FP1 spec (just one teeny bit of content, but that might be what caused the mark drop). However, F1/FP1 papers are not inherently harder than one another, one may be harder iin a particular year as a coincidence, but it is not intended.
    Nah IAL papers are defo harder tbf
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    (Original post by BBeyond)
    Nah IAL papers are defo harder tbf
    yh the jan or june 14 for s2 was the only one where it was p1ss easy incomparison to the bog standard.

    Tho the F1 papers seem harder then ial which are harder than the r papers which are harder than the normal papers imo

    annoyingly boundaries remain the same so a hard paper means you still need a high mark for the 90+ ums (hoping this isn;t the case in 2 weeks lol)
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    Could someone explain the last 2 terms for me please?
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    (Original post by Chirstos Ioannou)
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    Could someone explain the last 2 terms for me please?
    \displaystyle \sum_{r=0}^{n} (2n+1) = (2n+1)\sum_{r=0}^{n} 1 = (2n+1)(\underbrace{1 + 1 + 1 + \cdots + 1}_{(n+1) \, \text{times}}) = (2n+1)(n+1)

    There are (n+1) 1's because the sum starts from 0.
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    (Original post by Zacken)
    \displaystyle \sum_{r=0}^{n} (2n+1) = (2n+1)\sum_{r=0}^{n} 1 = (2n+1)(\underbrace{1 + 1 + 1 + \cdots + 1}_{(n+1) \, \text{times}}) = (2n+1)(n+1)

    There are (n+1) 1's because the sum starts from 0.
    So the reason you are able to pull (2n+1) out of the sum expression is because you can think of it as a number and not a variable?
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    (Original post by Chirstos Ioannou)
    So the reason you are able to pull (2n+1) out of the sum expression is because you can think of it as a number and not a variable?
    It's not a variable. You can't pull 'r' out because you're summing over r.

    Basically, in a sum \sum_{r=0}^{n} you can pull out anything except terms that contain the letter that is at the bottom of the sum, so in this case, you can't pull out r, but you can pull out anything else.

    If you had \sum_{k=0}^{n} you can pull out anything except k, etc...
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    (Original post by Zacken)
    It's not a variable. You can't pull 'r' out because you're summing over r.

    Basically, in a sum \sum_{r=0}^{n} you can pull out anything except terms that contain the letter that is at the bottom of the sum, so in this case, you can't pull out r, but you can pull out anything else.

    If you had \sum_{k=0}^{n} you can pull out anything except k, etc...
    Yeap got it, thanks
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    (Original post by Chirstos Ioannou)
    Yeap got it, thanks
    No worries.
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    Just came over to this thread, I did the exam last year so ask away!
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    But why are the first two terms the same as in the formula since the sum starts from r=0?
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    (Original post by Zacken)
    No worries.
    But why are the first two terms the same as in the formula since the sum starts from r=0?
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    (Original post by Chirstos Ioannou)
    But why are the first two terms the same as in the formula since the sum starts from r=0?
    Because you can split them up like so:

    \displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

    i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).
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    (Original post by Zacken)
    Because you can split them up like so:

    \displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

    i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).
    Exactly
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    (Original post by Zacken)
    Because you can split them up like so:

    \displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

    i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).
    So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?
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    (Original post by Redcoats)
    Exactly
    ?

    (Original post by Chirstos Ioannou)
    So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?
    Yeah.
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    (Original post by Zacken)
    ?



    Yeah.
    Ok thank you

    He just agreed to your explanation.
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    (Original post by Chirstos Ioannou)
    So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?
    Yes - it is really useful to write out the sequence; I did that all over my FP1 exam!
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    (Original post by Chirstos Ioannou)
    Ok thank you

    He just agreed to your explanation.
    Yeah, if you find yourself getting confused by the sigma notation, write out the series explicitly.
 
 
 
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