Question

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.

Answer 1

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.