Section8.4An Application of the Binomial Theorem¶ permalink
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.
Lemma8.10
For each \(k\ge0\),
\(P(p,k+1)=P(p,k)(p-k)\).
Proof
When \(k=0\), both sides evaluate to \(p\). Now assume validity when
\(k=m\) for some non-negative integer \(m\). Then
\begin{align*}
P(p,m+2)\amp =pP(p-1,m+1)\\
\amp = p[P(p-1,m)(p-1-m)]\\
\amp =[pP(p-1,m)](p-1-m)\\
\amp =P(p,m+1)[p-(m+1)].
\end{align*}
Our goal in this section will be to invoke Newton's Binomial Theorem with the exponent \(p=-1/2\). To do so in a meaningful manner, we need a simplified expression for \(C(-1/2,k)\), which the next lemma provides.
Lemma8.11
For each \(k\ge0\),
\(\displaystyle\binom{-1/2}{k}=(-1)^k\frac{\binom{2k}{k}}{2^{2k}}\).
Proof
We proceed by induction on \(k\). Both sides reduce to \(1\) when \(k=0\). Now assume validity when \(k=m\) for some non-negative integer \(m\). Then
\begin{align*}
\binom{-1/2}{m+1} \amp =\frac{P(-1/2,m+1)}{(m+1)!}
=\frac{P(-1/2,m)(-1/2-m)}{(m+1)m!}\\
\amp =\frac{-1/2-m}{m+1}\binom{-1/2}{m}
=(-1)\frac{2m+1}{2(m+1)}(-1)^m\frac{\binom{2m}{m}}{2^{2m}}\\
\amp =(-1)^{m+1}\frac{1}{2^{2m}}
\frac{(2m+2)(2m+1)}{(2m+2)2(m+1)}\binom{2m}{m}=
(-1)^{m+1}\frac{\binom{2m+2}{m+2}}{2^{2m+2}}.
\end{align*}
Theorem8.12
The function \(f(x)=(1-4x)^{-1/2}\) is the generating function of the sequence \(\{\binom{2n}{n}:n\ge0\}\).
Proof
By Newton's Binomial Theorem and Lemma 8.11, we know that
\begin{align*}
(1-4x)^{-1/2}\amp =\sum_{n=0}^\infty\binom{-1/2}{n}(-4x)^n\\
\amp =\sum_{n=0}^\infty(-1)^n2^{2n}\binom{-1/2}{n}x^n\\
\amp =\sum_{n=0}^\infty \binom{2n}{n}x^n.
\end{align*}
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 \(f(x) = (1-4x)^{-1/2}\).
Corollary8.13
For all \(n\ge0\),
\begin{equation*}
2^{2n}=\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{k}.
\end{equation*}