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    (Original post by Heroesorghosts)
    I understand the general concept i.e. If g(r)=f(r+1)-f(r) for r\in\mathbb{N}, then \sum_{r=1}^n g(r) =f(n+1)-f(1), however I can't seem to be able to apply it to questions? I tend to go down the route of simply writing out the first few lines of the summing, then skipping to n and seeing which cancel out and which don't! For example in the following question,

    Verify that \frac{1}{r}-\frac{1}{r+1}=\frac{1}{r(r+1)}. Hence show that \sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}.

    I can do the first part fine, however in the second part i.e. using the difference method above, the example defines g(r)= \frac{1}{r(r+1)} and f(r)=-\frac{1}{r}. The part I don't understand is changing the f(r) to negative?! My teacher explained it so that the final equation i.e. when you put it in the \sum_{r=1}^n g(r) =f(n+1)-f(1) formula, it becomes positive again, but I still don't fully understand this? Also, if anyone could give me a few pointers on the following question, it would be greatly appreciated. I am fine on the first part, it's just using the difference method again that gets me!

    \frac{\frac{1}{2}}{r-1}-\frac{\frac{1}{2}}{r+1}=\frac{1}  {r^2-1}. Show that \sum_{r=2}^n \frac{1}{r^2-1} = \frac{3}{4}-\frac{2n+1}{2n(n+1)}. Thanks
    The reason if is f(r)=-1/r is because essentially, your formula is the "wrong way round" notice in general you want f(r+1)-f(r) but you have in this case f(r)-f(r+1) essentially, sticking a minus sign in front of f reverses it to the one you want for the general formula which is why they chose this particular f
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    I'm not sure how they want you to go about the second part, but here's how I'd do it: consider g(r)=f(r+1)-f(r) and g(r)=f(r)-f(r-1). Add these together gives 2g(r)=f(r+1)-f(r-1) or g(r)=(f(r+1)-f(r-1))/2. Do a similar thing with the sum and then just figure out what f is (note: it will be negative as in your example for the first question because of the reversal of the signs)
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    (Original post by Heroesorghosts)
    I understand the general concept i.e. If g(r)=f(r+1)-f(r) for r\in\mathbb{N}, then \sum_{r=1}^n g(r) =f(n+1)-f(1), however I can't seem to be able to apply it to questions? I tend to go down the route of simply writing out the first few lines of the summing, then skipping to n and seeing which cancel out and which don't!
    This is exactly the way to do it and also how you should write your working out, even in the exam. If asked to sum 1/r - 1/(r+1) from 1 to n, I would literally write this and expect full marks:

    1/1 - 1/2
    + 1/2 - 1/3
    + ...
    + 1/(n-1) - 1/n
    + 1/n - 1/(n+1)

    = 1 - 1/(n+1) = n/(n+1).

    (cross out the terms that cancel on paper as well)

    As for your second question, \sum \frac{1}{r^2 - 1} = 0.5 \sum (\frac{1}{r-1} - \frac{1}{r+1})

    = 0.5(1/1 - 1/3
    + 1/2 - 1/4
    + 1/3 - 1/5
    + 1/4 - 1/6
    + 1/5 - 1/7
    ...
    + 1/(n-1) - 1/(n+1))

    Can you see how it'll work? Up to where I've let r = 6, cross out what cancels and then think about which terms remain in terms of n (e.g. the last term on the right stay, so that's -1/(n+1), the second last term on the right stays, so that's -1/n, etc).
 
 
 
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