Binomial expansion
posted by Helga .
The coefficient of x^2 in the expansion of (43x)^n as a series of ascending powers of x is 9/64. Show that n satisfies the equation 4^(n+1)=2/[n(1n)] and hence verify that n=1/2.
I can do the first part of the problem:
4^n*3^2*n(n1)/2!*2^4=3^2/2^6
4*4^n*n(n1)=2
4^(n+1)*n(n1)=2
4^(n+1)=2/[n(1n)]
But unfortunately I'm stuck with the verification of n=1/2.
Please help.

The binomial expansion is usually expressed as:
(p+q)^n
=∑ (n,r)p^i*q^(ni)
for i=0,n
where (n,r)=n!/[r!(nr)!]
so for
p=4
q=3x
i=(n2)
(n,r)=n*(n1)/2!
[note: (n,r)=(n,nr), so (n,n2)=(n,2)]
so term n2 is
[n(n1)/2!]*[4^(n2)]*[(3x)^2]
=[n(n1)/2!]*[4^(n2)]*[9]x²
and the coefficient
[n(n1)/2!]*[4^(n2)]*[9] = 9/64
Solve for n by trying various values of n that gives 9/64 on the lefthand side.
I get n=1/2.