I am trying to solve assignment of an Institute in which I don't study and I am unable to get answer of this question. For theory, I am studying Introductory Combinatorics by Richard Brualdi.

Question is -> Show that $ (1-4x)^{-1/2} $ generates the sequence $ {2n \choose n} $ .

Using Newton Binomial formula I am getting $\frac{ (2n) (2n-2) ... (2n-2k +2) 2^k } { k! } $ which will not yield the answer.

I don't know where I am mistaken . Can someone please give a detained answer. I cannot ask anyone as I am self studying.


You obtained something with two variables: $n$ and $k$. The required coefficient depends on $n$ only!

Using the (generalized) binomial formula means that $$(1-4x)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}(-4)^{n} x^n.$$ Now expand the binomial coefficient $$\begin{align} \binom{-1/2}{n}(-4)^{n}&=\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})\dots (-\frac{1}{2}-n+1)(-4)^{n}}{n!}\\ &=\frac{(1)(3)(5)\dots (2n-1)(2)^{n}}{n!} \end{align}$$ Can you take it from here?


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