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    (Original post by Zacken)
    There's no infinity involved here. If I'm using k then that's any arbitrary integer that does not imply infinity in anyway.

    Everything cancels out, but remember that in the \cdots there's something that cancels the 2k-3 as well, so you're only left with...?
    1/(2k + 1)?
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    (Original post by TheOtherSide.)
    1/(2k + 1)?
    That's the denominator, yes. So what's 1/that?
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    (Original post by Zacken)
    That's the denominator, yes. So what's 1/that?
    I have got no idea what to do with this :facepalm:
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    (Original post by TheOtherSide.)
    I have got no idea what to do with this :facepalm:
    So you've simplified \frac{1}{\text{blah}} to \displaystyle \frac{1}{\frac{1}{2k+1}} can you simplify this any further?
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    (Original post by Zacken)
    So you've simplified \frac{1}{\text{blah}} to \displaystyle \frac{1}{\frac{1}{2k+1}} can you simplify this any further?
    Oh right, that's the same as just 2k + 1.
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    (Original post by TheOtherSide.)
    Oh right, that's the same as just 2k + 1.
    And that's the answer. :yep:
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    (Original post by Zacken)
    And that's the answer. :yep:
    Thanks for the help! :woo:
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    (Original post by TheOtherSide.)
    Thanks for the help! :woo:
    Good work!
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    (Original post by notnek)
    I sitll find that reaction crazy. It's a standard A* question and similar to ones I've seen in old GCSE and IGCSE papers.

    If any GCSE students want a challenging paper, take a look at the Edexcel June 2004 non-calc. The A* boundary was only 61%.
    Saw your post so I tried to complete it. Managed to get 88% but damn... so much harder than usual
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    (Original post by surina16)
    Saw your post so I tried to complete it. Managed to get 88% but damn... so much harder than usual
    Yes I find that top students who get close to 100% in most GCSE past papers often get below 90 in the June 04 paper.
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    A circle is inscribed in an equilateral triangle of side 2, so that the sides of the triangle are tangent to the circle. Find the ratio of the area of the circle to the area of the triangle.
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    Find the value of

    \displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1  + \cdots}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
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    (Original post by atsruser)
    Find the value of

    \displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1  + \cdots}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
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    Find the value of

    \displaystyle \frac{1}{1+\frac{2}{1+\frac{1}{1  + \frac{2}{1+\cdots}}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
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    (Original post by atsruser)
    Find the value of

    \displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1  + \cdots}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
    Solution:
    Spoiler:
    Show
    Call it y, then \displaystyle y = \frac{1}{1 +y} \Rightarrow y^2 + y - 1 = 0 \Rightarrow y = \frac{1}{2}\left(\sqrt{5} - 1\right) where we pick the positive root because the fraction looks positive. Sorry for being so hand-wavey.


    What's with the recursive mood over the past few days? :lol:
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    (Original post by atsruser)
    Find the value of

    \displaystyle \frac{1}{1+\frac{2}{1+\frac{1}{1  + \frac{2}{1+\cdots}}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
    Spoiler:
    Show
    Call it y then \displaystyle y = \frac{1}{1 + \frac{2}{1 + y}} = \frac{1+y}{3+y} \Rightarrow y^2 + 2y - 1 = 0 \Rightarrow y = \sqrt{2} - 1.


    D'you have a better way to justify taking the positive root?
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    (Original post by atsruser)
    Find the value of

    \displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1  + \cdots}}}

    where \cdots indicates that the fraction continues the same pattern to infinity.
    And for a bonus mark(or rather, for a gold star :ahee:), what is the special name given to 1 + \displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1  + \cdots}}}
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    (Original post by atsruser)
    You're cheating as you're not a GCSE student. You really ought to spoiler these.
    Apologies. I've spoilered them, might want to edit your post so you can remove the quote.
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    (Original post by Zacken)
    Apologies. I've spoilered them, might want to edit your post so you can remove the quote.
    No need to apologise - I find it hard to stop myself solving problems if I can see how to do them.
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    (Original post by Zacken)
    This 6 mark question came up in my GCSE exam and I quite enjoyed it.Integers a and b are such that (a + 3\sqrt{5})^2 + a - b\sqrt{5} = 51. Find the possible values of a and the corresponding values of b.
    how do you do this? I expanded the double brackets and got

    a^2+6 sqrt(5) a+45
 
 
 
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