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    So my weakness, rearranging stuff comes back to haunt me again...
    Q2 is : Find the value of n for which U_n has the given value:
    U_n= (-1)^n \frac{n}{n+4}, U_n= \frac{7}{9}

    So from that, I have

    (-1)^n \frac{n}{n+4}= \frac{7}{9}

    and don't know where to go from there. I'm guessing it involves multiplying out the n+4 but don't really know
    Help please, oh lovely mathemeticians? :o: ty x
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    I've never seen C1 questions written in Latex before but I'm suitably impressed.

    Notice you want to rewrite 7/9 in a form where the denominator is 4 greater than the numerator, i.e. write 7/9=14/18 which should make the answer clear.
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    That's a nice method from Gaz there, use that ^^^

    Alternatively, you could do some basic case analysis. Firstly make the assumption that the value of n you're trying to find is even. See if that assumption leads to a solution, but remember you assumed n is even so the only valid solutions to this equation will be even. Then do exactly the same thing except assume that n is odd.

    The idea is that by doing this you can get rid of the (-1)^n. If n is off then it's always equal to -1 whereas if n is even it's always equal to 1. In either case you should have a relatively easy equation to solve.

    But like I say, Gaz's method is much nicer. Use that
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    Thanks for your help, but I'm afraid I don't get it, the dummy I am

    So now I've got \frac {(-1)^n n}{\frac{14}{18}} = n+4

    I doubt that's right but if it is, I'm stuck again haha. Sorry :P
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    I think he meant that

    \frac{14}{18} = \frac{n(-1)^n}{n+4}

    You can just read off n from this. It's a clever trick but I think JohnnySPal's method is better for your knowledge.

    *Edit*

    If you're still not following then:

    \frac{14}{14 + 4} = \frac{n(-1)^n}{n+4}
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    Ooo.. ty :]
 
 
 
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Updated: November 15, 2008

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