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    edexcel p2 jan 02 q 9

    f(x)=(1+x/k)n n>2 k,n E N

    the coefficient of x3 is twice the coefficient of x2 in the binomail explansion of f(x)

    prove that n=6k+2

    so I said

    2 n!/2(n-2)!k2 =n!/6(n-3)(n-2)!k3

    cancelling out gives

    1/k2 =1/6(n-3)k3

    then 1=(6n-18)k

    then I get stuck and cant rearrange to get n=


    Also the mark scheem is very different

    they give

    2n(n-1)/2k2 =n(n-1)(n-2)/6k3

    6k=n-2
    n=6k+2

    help please
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    2 n!/2(n-2)!k2 =n!/6(n-3)(n-2)!k3
    This is wrong. ^^

    The correct expansion is:

    (1 + x/k)n = 1 + n(x/k) + [n(n-1)(x/k)2]/2 + [n(n-1)(n-2)(x/k)3/6 ...

    So: 2[n(n-1)(1/k)2]/2 = [n(n-1)(n-2)(1/k)3/6

    => 2/2 = (n - 2)/6k

    => 6k = n - 2

    => n = 6k + 2
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    Yeah i kinda dont understand why ur using factorials everywhere??

    f(x)=(1+x/k)n

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


    if 2x² = x³ coeff then
    2(n(n-1)(1/k)²)/(2!) = (n(n-1)(n-2)(1/k)³)/(3!)
    1/2k²= (n-2)/6k³
    6k = n-2
    n=6k+2
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    (Original post by samdavyson)
    This is wrong. ^^

    The correct expansion is:

    (1 + x/k)n = 1 + n(x/k) + [n(n-1)(x/k)2]/2 + [n(n-1)(n-2)(x/k)3/6 ...

    So: 2[n(n-1)(1/k)2]/2 = [n(n-1)(n-2)(1/k)3/6

    => 2/2 = (n - 2)/6k

    => 6k = n - 2

    => n = 6k + 2
    where does the correct expansion come from, i don't understand where the (n-1) come from.

    I was taught to solve using factorials to change n c r into n!/r!(n-r)!

    so they come from the nc2 and nc3
 
 
 
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