Prove that: $r^{\underline{k}}\:\cdot \:\left(r-\frac{1}{2}\right)^{\underline{k}}\:=\:\frac{\left(2r\right)^{\underline{2k}}}{2^{2k}}$

Question

I have to prove for $r \in \mathbb R$ and $k \in \mathbb N$, that:

$r^{\underline{k}}\:\cdot \:\left(r-\frac{1}{2}\right)^{\underline{k}}\:=\:\frac{\left(2r\right)^{\underline{2k}}}{2^{2k}}$

I tried resolving it by two ways but i got stuck. (I) Induction (II) Ordinary Factorial expression

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(I)

Base case $k\:=\:0$, $\:\:\:r^{\underline{0}}\:\cdot \:\left(r-\frac{1}{2}\right)^{\underline{0}}\:=\:\frac{\left(2r\right)^{\underline{2*0}}}{2^{2*0}}\:=1$

Hypothesis: $k=p,\:\:\:r^{\underline{p}}\:\cdot \:\:\left(r-\frac{1}{2}\right)^{\underline{p}}\:=\:\frac{\left(2r\right)^{\underline{2p}}}{2^{2p}}$

Thesis: $k=p+1,\:\:\:r^{\underline{p+1}}\:\cdot \:\:\left(r-\frac{1}{2}\right)^{\underline{p+1}}\:=\:\frac{\left(2r\right)^{\underline{2(p+1)}}}{2^{2(p+1)}}$

**Apply the fundamental property

$r^{\underline{p}}\:\left(r-p\right)^{\underline{1}}\:\left(r-\frac{1}{2}\right)^{\underline{p}}\:\left(r-\frac{1}{2}-p\right)^{\underline{1}}\:=\:\frac{\left(2r\right)^{\underline{2(p+1)}}}{2^{2\left(p+1\right)}}$

$r^{\underline{p}}\:\left(r-p\right)\:\left(r-\frac{1}{2}\right)^{\underline{p}}\:\left(r-\frac{1}{2}-p\right)\:=\:\frac{\left(2r\right)^{\underline{2(p+1)}}}{2^{2\left(p+1\right)}}$

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(II)

$r^{\underline{k}}\:\cdot \:\left(r-\frac{1}{2}\right)^{\underline{k}}\:=\:\frac{\left(2r\right)^{\underline{2k}}}{2^{2k}}$

$\frac{r!}{\left(r-k\right)!}\cdot \frac{\left(r-\frac{1}{2}\right)!}{\left(\left(r-\frac{1}{2}\right)-k\right)!}=\frac{\left(2r\right)!}{\left(2r-2k\right)!}\cdot \frac{1}{2^{2k}}$

$\frac{r!\frac{2r-1}{2}!}{\left(r-k\right)!\frac{2r-1-2k}{2}!}=\frac{\left(2r\right)!}{\left(2r-2k\right)4^k}$

Answer 1

We have $$ r^{\underline{k}}\left(r- \frac12 \right)^{\underline{k}} =\prod_{i=r-k+1}^r i \left(i-\frac12 \right)=2^{-2k}\prod_{i=r-k+1}^r 2i (2i-1) = 2^{-2k}(2r)^{\underline{2k}} .$$ Let me know if any step is unclear.

EDIT: Induction also works, I give here just the inductive step: $$ r^{\underline{k+1}}\left(r- \frac12 \right)^{\underline{k+1}} = r^{\underline{k}}\left(r- \frac12 \right)^{\underline{k}} (r-k)\left(r-k-\frac12\right) = \frac{(2r)^{\underline{2k}}}{2^{2k}} (r-k)\left(r-k-\frac12 \right) .$$ The last step uses the induction hypothesis. Now, $$ \frac{(2r)^{\underline{2k}}}{2^{2k}} (r-k)\left(r-k-\frac12 \right) = \frac{(2r)^{\underline{2k}}}{2^{2(k+1)}} (2r-2k)(2r-2k-1) = \frac{(2r)^{\underline{2(k+1)}}}{2^{2(k+1)}}. $$

Answer 2

By definition $$ (m)^{\underline{n}}=\frac{m!}{(m-n)!}=\frac{\Gamma(m+1)}{\Gamma(m-n+1)}; $$ hence, $$ (2r)^{\underline{2k}}=\frac{\Gamma(2(r+1/2))}{\Gamma(2(r-k+1/2))}. $$ From the Gamma duplication formula $$ \begin{align} (2r)^{\underline{2k}} &=\frac{\Gamma(r+1/2)\Gamma(r+1)}{2^{1-(2r+1)}\sqrt\pi} \frac{2^{1-(2r-2k+1)}\sqrt\pi}{\Gamma(r-k+1/2)\Gamma(r-k+1)}\\ &=2^{-2k}\frac{\Gamma(r+1/2)\Gamma(r+1)}{\Gamma(r-k+1/2)\Gamma(r-k+1)}\\ &=2^{-2k}r^{\underline{k}}\frac{\Gamma(r-1/2+1)}{\Gamma(r-1/2-k+1)}\\ &=2^{-2k}r^{\underline{k}}(r-1/2)^{\underline{k}}, \end{align} $$ which is the desired result.