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December 20, 2014

December 20, 2014

Posted by **Helga** on Wednesday, February 29, 2012 at 8:41am.

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 -
**MathMate**, Wednesday, February 29, 2012 at 10:04amThe 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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