Lemma8.10
For each \(k\ge0\text{,}\) \(P(p,k+1)=P(p,k)(p-k)\text{.}\)
In this section, we see how Newton's Binomial Theorem can be used to derive another useful identity. We begin by establishing a different recursive formula for \(P(p,k)\) than was used in our definition of it.
For each \(k\ge0\text{,}\) \(P(p,k+1)=P(p,k)(p-k)\text{.}\)
Our goal in this section will be to invoke Newton's Binomial Theorem with the exponent \(p=-1/2\text{.}\) To do so in a meaningful manner, we need a simplified expression for \(C(-1/2,k)\text{,}\) which the next lemma provides.
For each \(k\ge0\text{,}\) \(\displaystyle\binom{-1/2}{k}=(-1)^k\frac{\binom{2k}{k}}{2^{2k}}\text{.}\)
The function \(f(x)=(1-4x)^{-1/2}\) is the generating function of the sequence \(\{\binom{2n}{n}:n\ge0\}\text{.}\)
Now recalling Proposition 8.3 about the coefficients in the product of two generating functions, we are able to deduce the following corollary of Theorem 8.12 by squaring the function \(f(x) = (1-4x)^{-1/2}\text{.}\)
For all \(n\ge0\text{,}\) \begin{equation*} 2^{2n}=\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k}. \end{equation*}