Show that
$\ds
^{2r} \! {\rm C}_r =\frac{1\times3\times\dots\times (2r-1)}{r!} \, \times 2^r
\;,
$
for \(r\ge1\,\).
Give the first four terms of the binomial series for
\(\l 1 - p \r^{-\frac12}\).
By choosing a suitable value for \(p\) in this series, or otherwise, show that
$$
\displaystyle \sum_{r=0}^\infty \frac{ {\vphantom {\A}}^{2r} \! {\rm C}_r }{ 8^r} = \sqrt 2
\;
.$$ Show that
$$
\displaystyle
\sum_{r=0}^\infty
\frac{\l 2r + 1 \r \; {\vphantom{A}}^{2r} \! {\rm C} _r }{ 5^r} =\big( \sqrt 5\big)^3
\;.
$$
[{\bf Note: }
$
{\vphantom{A}}^n {\rm C}_r
$
is an alternative notation for
$\ds \
\binom n r
\,
\( for \)r\ge1\,\(, and \)
{\vphantom{A}}^0 {\rm C}_0 =1
$ .]
Show Solution
Solution:
\begin{align*}
\binom{2r}{r} &= \frac{(2r)!}{r!r!} \\
&= \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2r-1)(2r)}{r! r!} \\
&= \frac{1 \cdot 3 \cdot 5 \cdots (2r-1) \cdot (2 \cdot 1) \cdot (2 \cdot 2) \cdots (2 \cdot r)}{r!}{r!} \\
&= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot 1 \cdot 2 \cdots r}{r!r!} \\
&= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot r!}{r!r!} \\
&= \frac{1\cdot 3 \cdots (2r-1)}{r!} \cdot 2^r
\end{align*}
which is what we wanted to show
\begin{align*}
(1 - p)^{-\frac12} &= 1 + \left ( -\frac12 \right )(-p) + \frac{1}{2!} \left (-\frac12 \right )\left (-\frac32 \right )(-p)^2 + \ldots \\
& \quad \quad \quad \cdots +\frac{1}{3!} \left (-\frac12 \right )\left (-\frac32 \right )\left (-\frac52 \right )(-p)^3 + O(p^4) \\
&= \boxed{1 + \frac{1}{2}p + \frac{3}{8}p^2 + \frac{5}{16}p^3} + O(p^4)
\end{align*}
More generally:
\begin{align*}
\binom{-\frac{1}{2}}{k} &=\frac{(-\frac{1}{2})\cdot(-\frac{1}{2} -1)\cdots(-\frac12 -k+1)}{k!} \\
&= \frac{(-1)(-3)(-5)\cdots(-(2k-1))}{k!2^k} \\
&= \frac{(-1)^k(1)(3)(5)\cdots((2k-1))}{k!2^k} \\
&= (-1)^k \frac{1}{4^k}\binom{2k}{k} \\
\end{align*}
Therefore,
\begin{align*}
\sqrt{2} = \left (1-\frac12 \right)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-\frac12 \right )^r \tag{\(\frac12 < 1\) so series is valid} \\
&= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} \left (-\frac12 \right )^r \\
&= \sum_{r=0}^{\infty} \frac{1}{8^r}\binom{2r}{r}
\end{align*}, which is what we wanted to show. \begin{align*}
p(1-p^2)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-p^2 \right )^rp \\
&= \sum_{r=0}^{\infty} \frac{1}{4^r}\binom{2r}{r} p^{2r+1}
\end{align*}
Differentiating with respect to \(p\),
\begin{align*}
(1-p^2)^{-\frac12} +p^2(1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \\
(1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r}
\end{align*}
Letting \(p = \frac{2}{\sqrt{5}}\), and \(|\frac2{\sqrt{5}}| < 1\) we have
\begin{align*}
\left (1-\frac45 \right )^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \\
(\sqrt{5})^3 &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r}
\end{align*}
(Alternative)
\begin{align*}
(\sqrt5)^3 &= \left ( \frac{1}{5} \right )^{-\frac32} \\
&= \left ( 1- \frac{4}{5} \right )^{-\frac32} \\
&= \sum_{r=0}^{\infty} \binom{-\frac32}{r} \left (-\frac45 \right)^r \\
&= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \frac{-\frac32-(r-1)}{-\frac12} \left (-\frac45 \right)^r \\
&= \sum_{r=0}^{\infty} \binom{-\frac12}{r} (2r+1) \left (-\frac45 \right)^r \\
&= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} (2r+1) \left (-\frac45 \right)^r \\
&= \sum_{r=0}^{\infty}(2r+1)\binom{2r}{r} \left (\frac15 \right)^r \\
(\sqrt{5})^3 &= \sum_{r=0}^{\infty}\frac{1}{5^r}(2r+1)\binom{2r}{r} \\
\end{align*}