Problem source

Show that 
 $\ds   
 ^{2r} \! {\rm C}_r =\frac{1\times3\times\dots\times (2r-1)}{r!} \, \times 2^r
 \;,
 $
 for $r\ge1\,$.
 
 \begin{questionparts}  
 \item 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 
 $$
 \displaystyle \sum_{r=0}^\infty \frac{  {\vphantom {\A}}^{2r} \! {\rm C}_r }{ 8^r} = \sqrt 2
 \;
 .$$  
  
 \item Show that 
 $$
 \displaystyle 
 \sum_{r=0}^\infty   
 \frac{\l 2r + 1 \r \; {\vphantom{A}}^{2r} \! {\rm C} _r }{ 5^r} =\big( \sqrt 5\big)^3
 \;.
 $$ 
 \end{questionparts}  
  
 [{\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
 $ .]

Solution source

\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{questionparts}
    \item \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.

    \item \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*}
\end{questionparts}