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The 2013 STEP Prep Thread!

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    (Original post by Maths_Lover)
    Good stuff.

    So - have you started looking at any STEP papers yet?
    As I mentioned before...
    Looked but only attempted one question :L


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by M2k13)
    As I mentioned before...
    Looked but only attempted one question :L


    This was posted from The Student Room's iPhone/iPad App
    I must have missed that post. :L

    You could try some more now that exams are over. Have you finished Core 4?
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    (Original post by wcp100)
    I'm spending my last FP3 lesson this term on STEP.
    Nice. I get to spend all of my remaining F.Maths lessons on STEP as the class have started M1 but I've done it.
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    this still doesnt beat the 2014 cambridge applicants thread page
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    (Original post by iCiaran)
    I'm with you for CompSci with maths , although I haven't looked at STEP at all yet, only just started C4 this week.
    Awesome, glad to find someone else wanting to do CompSci! And I think most people haven't even thought about STEP yet, so you'll still be ahead of the majority
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    (Original post by Rahul.S)
    this still doesnt beat the 2014 cambridge applicants thread page
    What the.....:lolwut: I mean this thread would be good for interview practice (I can justify anything)
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    (Original post by Maths_Lover)
    Nice. I get to spend all of my remaining F.Maths lessons on STEP as the class have started M1 but I've done it.
    I think I'm only going to do STEP I
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    (Original post by wcp100)
    I think I'm only going to do STEP I
    OK. How come you might not do the others?
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    (Original post by Maths_Lover)
    OK. How come you might not do the others?
    Too much stress considering I don't need them.
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    (Original post by wcp100)
    Too much stress considering I don't need them.
    Uh huh. That would be the most sensible idea.
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    (Original post by Maths_Lover)
    Uh huh. That would be the most sensible idea.
    Simplify:

     \displaystyle (1-x)\prod_{r=0}^n (1+x^{2^r})
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    (Original post by wcp100)
    Simplify:

     \displaystyle (1-x)\prod_{r=0}^n (1+x^{2^r})
    Spoiler:
    Show

     \displaystyle (1-x)\prod_{r=0}^n (1+x^{2^r})

    \displaystyle = (1-x) \left[(1+x^0)(1+x^2)(1+x^4) \cdots (1+x^{2^r}) \right]

    \displaystyle = 2(1-x)\left[1 + x^2 + x^4 + x^6 + x^8 +\cdots + x^{ \displaystyle\sum_{i=1}^n 2^i}\right] ,

    \displaystyle = 2(1-x)\left[1 + x^2 + x^4 + x^6 + x^8 +\cdots + x^{2(2^n-1)}\right]

    \displaystyle = 2(1-x)\left[\frac{x^{2^n}-1}{x^2-1} \right]

    \displaystyle = 2(1-x)\left[\frac{1-x^{2^n}}{1-x^2} \right]

    \displaystyle = 2(1-x)\left[\frac{1-x^{2^n}}{(1-x)(1+x)} \right]

    \displaystyle = \frac{2(1-x^{2^n})}{1+x}



    Sorry for the delay - I was watching anime.
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    Hey everyone. I want to start doing some STEP work for the exams next year so that i am thoroughly prepared. Where would be a good place to start? Should i just go over past papers, because the past papers are really difficult and overwhelming when i look at them. If you've done STEP, how did you find them in the end, and where did you start?
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    (Original post by Maths_Lover)
     \displaystyle (1-x)\prod_{r=0}^n (1+x^{2^r})

    \displaystyle = (1-x) \left[(1+x^0)(1+x^2)(1+x^4) \cdots (1+x^{2^r}) \right]

    \displaystyle = 2(1-x)\left[1 + x^2 + x^4 + x^6 + x^8 +\cdots + x^{ \displaystyle\sum_{i=1}^n 2^i}\right] ,

    \displaystyle = 2(1-x)\left[1 + x^2 + x^4 + x^6 + x^8 +\cdots + x^{2(2^n-1)}\right]

    \displaystyle = 2(1-x)\left[\frac{x^{2^n}-1}{x^2-1} \right]

    \displaystyle = 2(1-x)\left[\frac{1-x^{2^n}}{1-x^2} \right]

    \displaystyle = 2(1-x)\left[\frac{1-x^{2^n}}{(1-x)(1+x)} \right]

    \displaystyle = \frac{2(1-x^{2^n})}{1+x}

    Sorry for the delay - I was watching anime.

    EDIT: Should I be spoilering this?
    Yes

    Spoiler:
    Show
     (1-x)(1+x)(1+x^2)...
     =(1-x^2)(1+x^2)(1+x^4).....(1+x^{2^n  })
     =(1-x^4)....(1+x^{2^n})
    Can you see where this is going?


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    I'm just posting here to keep tabs on it and to say that if any wants any advice feel free to message me


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    (Original post by wcp100)
    Yes
    That's the easy bit. What about the next part?
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    (Original post by jack.hadamard)
    That's the easy bit. What about the next part?
    Yes it's easy. But it's brilliant.

    I've done that too I'm slightly stuck and busy though. It's the and hence that bit.
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    (Original post by wcp100)
    I've done that too I'm slightly stuck and busy though.
    I can give you a hint, if you tell me on which part you are stuck.

    EDIT: You might also want to keep this paper for practice before the exam!
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    (Original post by jack.hadamard)
    I can give you a hint, if you tell me on which part you are stuck.

    EDIT: You might also want to keep this paper for practice before the exam!
    I'm not doing STEP III officially.
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    (Original post by wcp100)
    Yes

    Spoiler:
    Show
     (1-x)(1+x)(1+x^2)...
     =(1-x^2)(1+x^2)(1+x^4).....(1+x^{2^n  })
     =(1-x^4)....(1+x^{2^n})
    Can you see where this is going?


    Woo. :woo:

    Spoiler:
    Show
    Indeed - difference of two squares all the way. Alternatively, you can spilt it up like this:

    (1-x)(1-x)\left[(1+x^2)(1+x^4)\cdots (1+x^{2^n})\right]

    \displaystyle =(1-x^2)\left[\frac{1-x^{2^n}}{1-x^2} \right] - as previously determined.

    = 1-x^{2^n}
Updated: June 12, 2013
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