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    Can someone please help me with this question my brain is just not functioning right now and I’m not getting anywhere with it Name:  BAD8BEB1-9F38-4E95-B4D4-21337A24B045.jpg.jpeg
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    Thank you
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    (Original post by Maariyas)
    Can someone please help me with this question my brain is just not functioning right now and I’m not getting anywhere with it Name:  BAD8BEB1-9F38-4E95-B4D4-21337A24B045.jpg.jpeg
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    Thank you
    Post your working so far.

    Have you tested it?

    Assume true for some n= k then look at n= k+1
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    it is more likely to be the method of differences
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    I got to the induction stage, my working is rough and a mess but this is where I got to p
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    Sorry didn’t attach it on the last one. kName:  46D1E6A1-4E30-4D9D-BF97-1F788B821B6E.jpg.jpeg
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    (Original post by the bear)
    it is more likely to be the method of differences
    What is the method of differences?
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    (Original post by Maariyas)
    Sorry didn’t attach it on the last one. k+3 should not be crossed on top and bottom
    This is fine. Now you want to fully factorise the numerator.
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    (Original post by RDKGames)
    This is fine. Now you want to fully factorise the numerator.
    i feel like I'm being really stupid but I'm just not seeing what the numerator would factorise into
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    (Original post by Maariyas)
    i feel like I'm being really stupid but I'm just not seeing what the numerator would factorise into
    You can get the fully expanded cubic, then guess a root (try factors of the constant term, and their -ve versions, like -1 and 1). Once you have one root, you can divide your cubic by the corresponding linear factor and obtain a quadratic. Then just factorise that quadratic to get the other 2 linear factors.
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    (Original post by RDKGames)
    You can get the fully expanded cubic, then guess a root (try factors of the constant term, and their -ve versions, like -1 and 1). Once you have one root, you can divide your cubic by the corresponding linear factor and obtain a quadratic. Then just factorise that quadratic to get the other 2 linear factors.
    but i thought i needed to collect like terms/common factors except i'm not seeing any
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    (Original post by Maariyas)
    but i thought i needed to collect like terms/common factors except i'm not seeing any
    Well, you can't expect ALL cases to be this simple.

    The numerator fully factorises, but not in this obvious way.
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    I’ve got it! Thank you
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    (Original post by Maariyas)
    I’ve got it! Thank you
    Well done.

    As to shed some light on what method of differences is, it's an alternative method to prove sums like these.

    Note that \dfrac{4}{r(r+2)}=\dfrac{2}{r}-\dfrac{2}{r+2} (check!) so then the sum is simply

    \begin{aligned} \sum_{r=1}^n \dfrac{2}{r}-\dfrac{2}{r+2} & = \left( \dfrac{2}{1} - \dfrac{2}{3} \right) + \left( \dfrac{2}{2} - \dfrac{2}{4} \right) + \left( \dfrac{2}{3} - \cancel{\dfrac{2}{5}} \right) + \left( \dfrac{2}{4} - \dfrac{2}{6} \right) + ... + \left( \dfrac{2}{n-1} - \dfrac{2}{n+1} \right) + \left( \dfrac{2}{n} - \dfrac{2}{n+2} \right) \\ & = \dfrac{2}{1}+\dfrac{2}{2}-\dfrac{2}{n+1} - \dfrac{2}{n+2} \\ & = 3-\dfrac{4n+6}{(n+1)(n+2)} \\ & = \dfrac{n(3n+5)}{(n+1)(n+2)}\end{  aligned}

    Second line is true because every other term cancels out within the sum (like -2/4 in second bracket is canceled with +2/4 in the 4th bracket, etc...)
 
 
 
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