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    Hi people,

    Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

    Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

    I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

    PS - there is rep at stake here!!
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    (Original post by Leekey)
    Hi people,

    Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

    Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

    I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

    PS - there is rep at stake here!!

    Rep at stake, here I come

    (1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

    Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-2 = 6, so n=8.
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    (Original post by Leekey)
    Hi people,

    Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

    Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

    I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

    PS - there is rep at stake here!!
    (1+x/2)^n = 1 + nx/2 + n(n-1)/2 x^2/4 + n(n-1)(n-2)/6 x^3 /8 + ...
    n(n-1)/8 = n(n-1)(n-2)/48

    1 = (n-2)/6

    n-2 = 6

    n=8
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    (Original post by theone)
    Rep at stake, here I come

    (1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

    Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-6 = 3, so n=9.
    but n=8
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    (Original post by elpaw)
    but n=8
    Check it
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    (Original post by theone)
    Check it
    even leekey said n=8

    where did you get n-6=3 from?
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    (Original post by elpaw)
    even leekey said n=8

    where did you get n-6=3 from?
    typo.
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    (Original post by theone)
    Rep at stake, here I come

    (1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

    Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-2 = 6, so n=8.
    WAA??
    I'm doing P2 and you've just lost me here mate..lol
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    (Original post by jndikum)
    WAA??
    I'm doing P2 and you've just lost me here mate..lol
    n>1 because otherwise you don't get a coeff of x^2 or x^3....and apart from that it's just binomial expansion...you'll come to it
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    (Original post by king of swords)
    n>1 because otherwise you don't get a coeff of x^2 or x^3....and apart from that it's just binomial expansion...you'll come to it
    I'm on the OCR syllabus..which are you on?
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    (Original post by theone)
    typo.
    crafty editing there. trying to win that rep, are you?
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    (Original post by jndikum)
    I'm on the OCR syllabus..which are you on?
    edexcel...remember, the evil examining board who made a mess up on a maths exam a few years back .
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    (Original post by elpaw)
    crafty editing there. trying to win that rep, are you?
    HAHAHA!! That's what I noticed.
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    (Original post by elpaw)
    crafty editing there. trying to win that rep, are you?
    Won't say no... and to be perfectly honest, I edited it straight after I posted it...
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    (Original post by theone)
    Won't say no... and to be perfectly honest, I edited it straight after I posted it...
    You and elpaw type too quick for me...i was going to get that rep .
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    (Original post by theone)
    Won't say no... and to be perfectly honest, I edited it straight after I posted it...
    straight after i pointed it out
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    (Original post by Adhsur)
    HAHAHA!! That's what I noticed.
    How dare you
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    so where is leekey to reap the rewards
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    (Original post by elpaw)
    so where is leekey to reap the rewards
    i think he tricked us all :confused:
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    (Original post by king of swords)
    i think he tricked us all :confused:
    as long as he doesnt come over all maskallesque and say he "already knew the answer hahahahahaha"
 
 
 
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