Deduce Taylor Series: (2n choose n) x^n Converges to 1/sqrt(1-4x)

[braindead101]

Deduce that the Taylor series about 0 of 1/sqrt(1-4x) is the series summation (2n choose n) x^n.
From this conclude that summation (2n choose n) x^n converges to 1/sqrt(1-4x) for x in (-1/4,1/4).
Then show that summation (2n choose n) (-1/4)^n = 1/sqrt(1-4(-1/4)) = 1/sqrt(2)

What I know:
Taylor series about 0 of f(x) = (1+x)^r, r is a real number given by summation (r choose n) x^n.
I know that (r choose n) can be rewritten as r(r-1)(r-2)..(r-n+1)/n!
and I know from a previous question that the Taylor series converges to f(x) for all x in (-1,1), and summation (2n choose n) x^n converges conditionally at x=-1/4.

How can I do this question with all this information? I am not sure how to piece it all together I am having a lot of trouble with this course.

 

[Pere Callahan]

Well, you know how to Taylor-expand the function \frac{1}{\sqrt{1+x}}=(1+x)^{-\frac{1}{2}}, it just your formula with r = -1/2.

So then just write -4x instead of x and you have the Taylor expansion for the given function

You might want to try and play around with the binomial coefficients ("r chose n") to bring it into the desired form.

 

[braindead101]

I have tried to do this, but I am stuck and cannot get them to equal each other.
For f(x)=(1+(-4x))^(-1/2)
The taylor series about 0 is:
sum (-1/2 choose n) (-4x)^n
expanding binomial coefficients:
sum -1/2(-1/2-1)(-1/2-2)...(-1/2-n+1) / n! x (-4)^n (x)^n
sum -1/2(-3/2)(-5/2)...(1/2-n)/n! x (-1)^n (4)^n (x)^n

While the series I am trying to get is:
sum (2n choose n) x^n
expanding binomial coefficients:
sum 2n(2n-1)(2n-2)..(2n-n+1)/n! (x)^n
sum 2n(n-1/2)(n-1)(n-3/2)...(n+1)/n! (x)^n

I am stuck now

 

[braindead101]

can anyone help me out, i am still stuck in same place