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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?
    They're not the same thing at all - r is a dummy variable, you could replace it with anything, i, j are frequently used as well. It basically signifies what variable in the equation you are incrementing each time.

    n however is the last number you would sum up to.

    The best way of explaining it is to make an analogy to integration:

    \displaystyle \int_1^n f(x) \, \mathrm{d}x

    Here x would be like r, a dummy variable - you increment x and then add, on and on, n is the limit of what you integrate/sum up to.

    So to sum up

    \displaystyle \sum_{r=1}^n f(n) = \sum_{i=1}^n f(n) = \sum_{j=1}^n f(n) = \cdots = F(n)

    However, \displaystyle \sum_{r=1}^n \neq \sum_{r=1}^m \neq \cdots \mathrm{etc...}
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    (Original post by physicsmaths)
    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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    Ah I see! Thanks!

    The other question I'm stuck on is finding (sigma) 2r+3/r(r+1) times by 1/3^r

    I've converted 2r+3/r(r+1) into partial fractions but I don't know what to convert 1/3^r to..
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    (Original post by creativebuzz)
    Ah I see! Thanks!

    The other question I'm stuck on is finding (sigma) 2r+3/r(r+1) times by 1/3^r

    I've converted 2r+3/r(r+1) into partial fractions but I don't know what to convert 1/3^r to..
    I'm not quite sure what the question is - would you mind posting a screenshot or something?
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    (Original post by Zacken)
    I'm not quite sure what the question is - would you mind posting a screenshot or something?
    Of course!
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    (Original post by creativebuzz)
    Of course!
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    You shouldn't be "converting" the 3^r to anything!

    On the face of it, that question looks quite horrible. THEREFORE you expect that applying partial fractions to the bit you can convert will simplify the sum to something that is going to be easy to evaluate! What do you get for your partial fraction expansion?
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    (Original post by davros)
    You shouldn't be "converting" the 3^r to anything!

    On the face of it, that question looks quite horrible. THEREFORE you expect that applying partial fractions to the bit you can convert will simplify the sum to something that is going to be easy to evaluate! What do you get for your partial fraction expansion?
    I got 3/r - 1/r+1

    but even if I did (sigma)2r+3/r(r+1) times by 1/3^r and do n=1, n=2 etc nothing really cancels out..
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    (Original post by creativebuzz)
    I got 3/r - 1/r+1

    but even if I did (sigma)2r+3/r(r+1) times by 1/3^r and do n=1, n=2 etc nothing really cancels out..
    Times that by 1/3^r the partial fraction part then sum it.


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    (Original post by creativebuzz)
    I got 3/r - 1/r+1

    but even if I did (sigma)2r+3/r(r+1) times by 1/3^r and do n=1, n=2 etc nothing really cancels out..
    I'm going offline shortly so can't check but it looks to me as if those partial fractions are going to be very helpful!

    Don't look for cancellations in the original form - look for cancellations when you write the result as two separate sums using the partial fractions
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    (Original post by creativebuzz)
    Of course!
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    You end up with \displaystyle \sum_{r=1}^n \left(\frac{3}{r} - \frac{1}{r+1}\right) \cdot 3^{-r} = \sum_{r=1}^n \left(\frac{3^{1-r}}{r} - \frac{3^{-r}}{r+1}\right)

    Which, if you write it out term by term, you see you get

    \displaystyle \left(1 - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{27}\right) + \left(\frac{1}{27} - \frac{1}{108}\right) + \cdots

    Can you take it from there?
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    (Original post by Zacken)
    You end up with \displaystyle \sum_{r=1}^n \left(\frac{3}{r} - \frac{1}{r+1}\right) \cdot 3^{-r} = \sum_{r=1}^n \left(\frac{3^{1-r}}{r} - \frac{3^{-r}}{r+1}\right)

    Which, if you write it out term by term, you see you get

    \displaystyle \left(1 - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{27}\right) + \left(\frac{1}{27} - \frac{1}{108}\right) + \cdots

    Can you take it from there?
    Ah yes, I'll give that a shot now!

    In the meantime, how would you go about drawing 3x+1/x-3 I found the asymptotes to be y=3 and x=3 but I don't know how to draw the shape of it..
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    (Original post by creativebuzz)
    Ah yes, I'll give that a shot now!

    In the meantime, how would you go about drawing 3x+1/x-3 I found the asymptotes to be y=3 and x=3 but I don't know how to draw the shape of it..
    I'm assuming you're talking about \frac{3x+1}{x-3} (use brackets to clear up ambiguity) - I would identify the asymptotes, as you have rightly done, then looked for x-intercepts (x = -\frac{1}{3}), y-intercepts, looked at what happens when x approaches 3 from the left or the right, maximum and minmum might be overkill but comes in useful every now and then, all of that combined should be enough to plot the graph.
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    (Original post by Zacken)
    I'm assuming you're talking about \frac{3x+1}{x-3} (use brackets to clear up ambiguity) - I would identify the asymptotes, as you have rightly done, then looked for x-intercepts (x = -\frac{1}{3}), y-intercepts, looked at what happens when x approaches 3 from the left or the right, maximum and minmum might be overkill but comes in useful every now and then, all of that combined should be enough to plot the graph.
    Yeah I've found the asymptotes and intercepts but I can't figure out what the shape of it would be still.. :/
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    (Original post by creativebuzz)
    Yeah I've found the asymptotes and intercepts but I can't figure out what the shape of it would be still.. :/
    Basically, you need to be staring at your graph with the asympotes drawn with dotted lines and your important points marked. Now, look at your vertical asymptote, what happens to the function as x gets closer to 3 from the left, that is, 2.95, 2.99, 2.999 - does the function get more negative, more positive, etc... In this case the function becomes increasingly negative (because you're dividing a positive number by a tiiiiiny negative number), so you know the curve should move downwards to infinity, but it can never touch the asymptote. Then, on the other side of the asymptote, your curve should start from the top of the graph instead because it suddenly became positive and it should move downwards gradually before levelling out at your y-asymptote. Can you understand?
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    (Original post by Zacken)
    Basically, you need to be staring at your graph with the asympotes drawn with dotted lines and your important points marked. Now, look at your vertical asymptote, what happens to the function as x gets closer to 3 from the left, that is, 2.95, 2.99, 2.999 - does the function get more negative, more positive, etc... In this case the function becomes increasingly negative (because you're dividing a positive number by a tiiiiiny negative number), so you know the curve should move downwards to infinity, but it can never touch the asymptote. Then, on the other side of the asymptote, your curve should start from the top of the graph instead because it suddenly became positive and it should move downwards gradually before levelling out at your y-asymptote. Can you understand?
    Ah yes, I think that made sense! If it was mod (3x+1)/x-3 would this be the correct graph?

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    (Original post by creativebuzz)
    Ah yes, I think that made sense! If it was mod (3x+1)/x-3 would this be the correct graph?

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    Not quuiiiiite. You're taking the modulus of a huge negative number, which would give you a huge positive number near x=3.

    Remember, modding a function just means reflecting what's under the x-axis in the x-axis. So your curve should curve upwards to x=3.
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    (Original post by Zacken)
    Not quuiiiiite. You're taking the modulus of a huge negative number, which would give you a huge positive number near x=3.

    Remember, modding a function just means reflecting what's under the x-axis in the x-axis. So your curve should curve upwards to x=3.
    But isn't that what I've done? I've flipped everything under the x axis so it's now above the x-axis
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    (Original post by creativebuzz)
    But isn't that what I've done? I've flipped everything under the x axis so it's now above the x-axis
    But when you flip something that's far below the x-axis, it should go far above the x-axis once you're reflected it.
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    (Original post by Zacken)
    But when you flip something that's far below the x-axis, it should go far above the x-axis once you're reflected it.
    Yeah, isn't that what I've done or is it not obvious enough?
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    (Original post by creativebuzz)
    Yeah, isn't that what I've done or is it not obvious enough?
    Ah, I see what you mean! But the graph still isn't right though, I don't see why it touches the x-axis and then shoots back up; it should look more like this:

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    Attachment 462099462101
    (Original post by Zacken)
    Ah, I see what you mean! But the graph still isn't right though, I don't see why it touches the x-axis and then shoots back up; it should look more like this:

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    Ah I see!

    Could you spot where I went wrong here?

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    (there are 2 attachments)

    The question was |5x+a|<=|2x|
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