Binomial expansion

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The coefficient of x^2 in the expansion of (4-3x)^n as a series of ascending powers of x is -9/64. Show that n satisfies the equation 4^(n+1)=2/[n(1-n)] and hence verify that n=1/2.

I can do the first part of the problem:
4^n*3^2*n(n-1)/2!*2^4=-3^2/2^6
4*4^n*n(n-1)=-2
4^(n+1)*n(n-1)=-2
4^(n+1)=2/[n(1-n)]

But unfortunately I'm stuck with the verification of n=1/2.
Please help.

  • Binomial expansion -

    The binomial expansion is usually expressed as:
    (p+q)^n
    =∑ (n,r)p^i*q^(n-i)
    for i=0,n
    where (n,r)=n!/[r!(n-r)!]

    so for
    p=4
    q=-3x
    i=(n-2)
    (n,r)=n*(n-1)/2!
    [note: (n,r)=(n,n-r), so (n,n-2)=(n,2)]
    so term n-2 is
    [n(n-1)/2!]*[4^(n-2)]*[(-3x)^2]
    =[n(n-1)/2!]*[4^(n-2)]*[9]x²
    and the coefficient
    [n(n-1)/2!]*[4^(n-2)]*[9] = -9/64
    Solve for n by trying various values of n that gives -9/64 on the left-hand side.

    I get n=1/2.

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