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    Could someone explain to me how to use the method of differences for summing series. i'm supposed to be self teaching it, but i don't really understand it.
    for example, could someone go through how i would find:
    \displaystyle\sum_{r=1}^n \frac{1}{2r(r+1)} - \frac{1}{2(r+1)(r+2)}

    thank you
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    An easy way to see what's going on is to write the first few terms of the series out, and see what's cancelling.
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    Substitute a few early values into it; r=1, r=2 and r=3 will usually suffice. That'll give each answer in two parts, the plus and minus parts. It should become apparent that the minus part from one will cancel out the plus part from another; keep substituting in the numbers until it becomes obvious which is cancelling out what (circling both of these might be helpful), e.g. each positive term cancels the negative term for the one that's 2 less than it, or something. Once that's apparent, then do a few later values; usually n-2, n-1 and n. Again, it'll become obvious which one's are cancelling out. This should leave a few values; there are those that you can ignore, as they'd be cancelled out if you'd just done some more values, and then some should be ones that you'll realise would never be cancelled out, and the sum is those terms that aren't cancelled out.
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    (Original post by Revolution is my Name)
    Substitute a few early values into it; r=1, r=2 and r=3 will usually suffice. That'll give each answer in two parts, the plus and minus parts. It should become apparent that the minus part from one will cancel out the plus part from another; keep substituting in the numbers until it becomes obvious which is cancelling out what (circling both of these might be helpful), e.g. each positive term cancels the negative term for the one that's 2 less than it, or something. Once that's apparent, then do a few later values; usually n-2, n-1 and n. Again, it'll become obvious which one's are cancelling out. This should leave a few values; there are those that you can ignore, as they'd be cancelled out if you'd just done some more values, and then some should be ones that you'll realise would never be cancelled out, and the sum is those terms that aren't cancelled out.
    okay so for my example one:

    r=1:  \frac{1}{4} - \frac{1}{12}

    r=2:  \frac{1}{12} - \frac{1}{24}

    r=3:  \frac{1}{24} - \frac{1}{40}

    r=n-2:  \frac{1}{2(n-2)(n-1)} - \frac{1}{2n(n-1)}

    r=n-1:  \frac{1}{2n(n-1)} - \frac{1}{2n(n+1)}

    r=n:  \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}

    so adding gives

     \frac{1}{4} - \frac{1}{12} +  \frac{1}{12} - \frac{1}{24} +  \frac{1}{24} - \frac{1}{40} + ... +  \frac{1}{2(n-2)(n-1)} - \frac{1}{2n(n-1)} +  \frac{1}{2n(n-1)} - \frac{1}{2n(n+1)} +  \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}

    which gives  \displaystyle\sum_{r=1}^n \frac{1}{2r(r+1)} - \frac{1}{2(r+1)(r+2)} = \frac{1}{4} - \frac{1}{40} + \frac{1}{2(n-2)(n-1)} - \frac{1}{2(n+1)(n+2)}
    is that right?
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    (Original post by apocalipse117)
    okay so for my example one:

    r=1:  \frac{1}{4} - \frac{1}{12}

    r=2:  \frac{1}{12} - \frac{1}{24}

    r=3:  \frac{1}{24} - \frac{1}{40}

    r=n-2:  \frac{1}{2(n-2)(n-1)} - \frac{1}{2n(n-1)}

    r=n-1:  \frac{1}{2n(n-1)} - \frac{1}{2n(n+1)}

    r=n:  \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}

    so adding gives

     \frac{1}{4} - \frac{1}{12} +  \frac{1}{12} - \frac{1}{24} +  \frac{1}{24} - \frac{1}{40} + ... +  \frac{1}{2(n-2)(n-1)} - \frac{1}{2n(n-1)} +  \frac{1}{2n(n-1)} - \frac{1}{2n(n+1)} +  \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}

    which gives  \displaystyle\sum_{r=1}^n \frac{1}{2r(r+1)} - \frac{1}{2(r+1)(r+2)} = \frac{1}{4} - \frac{1}{40} + \frac{1}{2(n-2)(n-1)} - \frac{1}{2(n+1)(n+2)}
    is that right?
    The 1/40 and 1/2(n-2)(n-1) terms will cancel out with things in the ... section of the sum.
    Can you see why?
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    (Original post by Slumpy)
    The 1/40 and 1/2(n-2)(n-1) terms will cancel out with things in the ... section of the sum.
    Can you see why?
    oh...erm possibly

    so if i do r=4:
    i get  \frac{1}{40} - \frac{1}{60}

    so the  \frac{1}{40} cancels and the  \frac{1}{60} would cancel with the first part of r=5

    oh wait, so then the \frac{1}{2(n-2)(n-1)}

    would cancel with the second part of r=n-3?

    so then the only terms not canceled are  \frac{1}{4} and  \frac{1}{2(n+1)(n+2)}

    so the answer would be  \frac{1}{4} + \frac{1}{2(n+1)(n+2)}?
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    (Original post by apocalipse117)
    oh...erm possibly

    so if i do r=4:
    i get  \frac{1}{40} - \frac{1}{60}

    so the  \frac{1}{40} cancels and the  \frac{1}{60} would cancel with the first part of r=5

    oh wait, so then the \frac{1}{2(n-2)(n-1)}

    would cancel with the second part of r=n-3?

    so then the only terms not canceled are  \frac{1}{4} and  \frac{1}{2(n+1)(n+2)}

    so the answer would be  \frac{1}{4} + \frac{1}{2(n+1)(n+2)}?
    Last term should be negative.
    Otherwise, yup.
 
 
 
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