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FP2: Method of Differences - Where have I gone wrong? watch

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    I've done part a and for part b I did the whole let n=1, n=2, n=3 blah blah and nothing cancelled out :/
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    (Original post by creativebuzz)
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    I've done part a and for part b I did the whole let n=1, n=2, n=3 blah blah and nothing cancelled out :/
    What did you get for n=1, n=2 and n=3?

    Think back to FP1 and this may simplify the problem.
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    (Original post by SeanFM)
    What did you get for n=1, n=2 and n=3?

    Think back to FP1 and this may simplify the problem.
    n=1:
    1/2

    n=2:
    7/6

    n=3:
    25/12

    Ah, I can't remember much of FP1 :/
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    (Original post by creativebuzz)
    n=1:
    1/2

    n=2:
    7/6

    n=3:
    25/12

    Ah, I can't remember much of FP1 :/
    Okay, don't worry

    Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?
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    (Original post by SeanFM)
    Okay, don't worry

    Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?

    Erm can you seperate them so you can have (sigma)r which is 1/2n(n+1) and (sigma)-1 which is just -1 (I think)
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    (Original post by SeanFM)
    Okay, don't worry

    Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?
    Why do you always torture people instead of giving the answer?
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    (Original post by creativebuzz)
    Erm can you seperate them so you can have (sigma)r which is 1/2n(n+1) and (sigma)-1 which is just -1 (I think)
    Almost

    Your first sum (n(n+1))/2 is right. Now think about the -1. If each 'r' contributes -1 to the overall sum.. how many -1's do you have? (In terms of n).
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    (Original post by mrno1324)
    Why do you always torture people instead of giving the answer?
    It'll help them to learn more, and giving the answer is against the spirit of this forum.
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    (Original post by SeanFM)
    Almost

    Your first sum (n(n+1))/2 is right. Now think about the -1. If each 'r' contributes -1 to the overall sum.. how many -1's do you have? (In terms of n).
    Hmm would you have r many -1s
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    (Original post by creativebuzz)
    Hmm would you have r many -1s
    Technically you'll have 'n' -1s but I know that that is what you meant. (As if you're summing the first 3 terms (n=3), when r=1 you get -1 from the -1, when r=2 you get another -1 all the way up to n.

    Now you can progress with the other half of the question
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    (Original post by SeanFM)
    Technically you'll have 'n' -1s but I know that that is what you meant. (As if you're summing the first 3 terms (n=3), when r=1 you get -1 from the -1, when r=2 you get another -1 all the way up to n.

    Now you can progress with the other half of the question
    But doesn't it just give you the same results

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    (Original post by creativebuzz)
    But doesn't it just give you the same results

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    Yes, if you just calculate without splitting up the terms.

    But think about why I've asked about the sum of r and -1 and what I mean by 'you're familiar with how to deal with (1/r)-(1/(r+1) when it's a sum'.
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    (Original post by SeanFM)
    Yes, if you just calculate without splitting up the terms.

    But think about why I've asked about the sum of r and -1 and what I mean by 'you're familiar with how to deal with (1/r)-(1/(r+1) when it's a sum'.
    Hmm I'm not entirely sure.. :/
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    (Original post by creativebuzz)
    Hmm I'm not entirely sure.. :/
    From part a you can see that the sum of that one fraction is the same as summing it up when it's broken down into four terms.

    The sum of all four terms can then be split up into the sum of the sum of each term.

    The sum of (1+2+3) is the same as the sum of (1) + the sum of (2) + the sum of (3).
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    (Original post by SeanFM)
    From part a you can see that the sum of that one fraction is the same as summing it up when it's broken down into four terms.

    The sum of all four terms can then be split up into the sum of the sum of each term.

    The sum of (1+2+3) is the same as the sum of (1) + the sum of (2) + the sum of (3).
    Yeah I understand that but I don't see how it's allowing my values to cancel out like they should
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    (Original post by creativebuzz)
    Yeah I understand that but I don't see how it's allowing my values to cancel out like they should
    Because you already know the sum of r and -1 for whatever n you choose (from FP1), then you're just concerned about (1/r) - (1/r+1), the sum of which you find by the method of differences.
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    (Original post by SeanFM)
    Because you already know the sum of r and -1 for whatever n you choose (from FP1), then you're just concerned about (1/r) - (1/r+1), the sum of which you find by the method of differences.
    Wait so n and r aren't the same thing?
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    (Original post by creativebuzz)
    Wait so n and r aren't the same thing?
    What SeanFM is saying might be slightly clearer to you by writing it out using eqations instead.

    He's saying that:

    \displaystyle \sum_{r=1}^n \left(r - 1 + \frac{1}{r} - \frac{1}{r-1}\right) = \sum_{r=1}^n r - \sum_{r=1}^n 1 + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

    This reduces to

    \displaystyle \frac{n(n+1)}{2} - n + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

    Now apply the method of differences to the last sum only, that is, sum up/evaluate the below sum individually.

    \displaystyle \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right) and then add that result to the first two terms in the above equation.
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    (Original post by Zacken)
    What SeanFM is saying might be slightly clearer to you by writing it out using eqations instead.

    He's saying that:

    \displaystyle \sum_{r=1}^n \left(r - 1 + \frac{1}{r} - \frac{1}{r-1}\right) = \sum_{r=1}^n r - \sum_{r=1}^n 1 + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

    This reduces to

    \displaystyle \frac{n(n+1)}{2} - n + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

    Now apply the method of differences to the last sum only, that is, sum up/evaluate the below sum individually.

    \displaystyle \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right) and then add that result to the first two terms in the above equation.
    Ah I see! But just to make clear, is n and r the same thing? If not, what's the difference?
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    (Original post by creativebuzz)
    Ah I see! But just to make clear, is n and r the same thing? If not, what's the difference?
    r represents what we are putting in r=1,2,3,4....,n-1,n
    For each term! Here it is represnting the general term and r takes all the values listed in the sum and n is one of them


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