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The Problem

Show that

$$\binom{1/2}{k} = \frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}$$

My Work

$$\begin{align*}\frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k} &= \frac{(-1)^{k+1}(2k)!}{4^k(2k-1)(k!)^2}\\\\ &= \frac{(-1)^{k+1}(2k)(2k-1)(2k-2)\cdots(k+1)}{4^k(2k-1)(k!)}\\\\ &= \frac{(-1)^{k+1}(2k)(2k-2)\cdots(k+1)}{4^k(k!)}\\\\ &= \frac{(-1)(2k)(2k-2)\cdots(k+1)}{(k!)}\times \left(\frac{-1}{4}\right)^k \end{align*}$$

I basically don't know where to go from here. My denominator is almost right where I want it to be, my numerator is fairly out of order though. Any suggestions on what I should try next, or if I should take a different approach all together?

Brian M. Scott
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Dunka
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1 Answers1

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HINT: It’s very often a good idea to play with both sides, trying to get them to meet in the middle. In this case it’s a good idea to begin expanding the lefthand side:

$$\begin{align*} \binom{1/2}k&=\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\ldots\left(-\frac{2k-3}2\right)}{k!}\\\\ &=\frac{(-1)^{k-1}(2k-3)(2k-5)(2k-7)\ldots(3)(1)}{2^kk!}\;.\tag{1} \end{align*}$$

Now go back and take another look at the result of your first step in manipulating the righthand side:

$$\frac{(-1)^{k+1}(2k)!}{4^k(2k-1)(k!)^2}\;.\tag{2}$$

Clearly $(-1)^{k+1}=(-1)^{k-1}$. The denominator of $(1)$ suggests splitting the denominator of $(2)$ as $(2^kk!)\cdot\big((2k-1)2^kk!\big)$ and trying to cancel the $(2k-1)2^kk!$ with something in the numerator. Thus, it looks as if you want to try to prove that

$$(2k-3)(2k-5)(2k-7)\ldots(3)(1)=\frac{(2k)!}{(2k-1)2^kk!}\;.$$

This isn’t too hard; try to combine $2^kk!$ in the denominator into a single product of $k$ integers.

Brian M. Scott
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  • Hmmm. Thanks for all the effort in this answer. Expanding the $(2k)!$ I see an easy cancellation for $(2k-1)$ there is also an easy cancellation for $k!$ but if I use that, I'm left without half of the LHS numerator and I still have a $2^k$ to take care of on the bottom – Dunka Feb 25 '15 at 03:09
  • @Dunka: You’re welcome. Don’t cancel the $k!$ in the obvious way: combine it with the $2^k$ first. You have $k$ factors of $2$, and $k$ factors in $k!$, ... – Brian M. Scott Feb 25 '15 at 03:11
  • I was trying to prove that $2^kk!$ would cancel out all the even cases in the denominator expansion with induction, then had a big "OOOOOO" moment as to why it made sense intuitively (anything(integers) multiplied by 2 is even. Thank you so much! – Dunka Feb 25 '15 at 03:36
  • @Dunka: Yep, you got it! You’re welcome! – Brian M. Scott Feb 25 '15 at 03:37