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  • How do I show that the product $$\biggl(4 - \frac21\biggr) \cdot \biggl(4 - \frac22\biggr) \cdot \biggl(4 - \frac23\biggr) \cdots \biggl(4 - \frac2{n}\biggr)$$ is an integer for any $n \in \mathbb{N}$.

Source: www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf

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Convert to a form involving the "double factorial": en.wikipedia.org/wiki/Factorial –  Blue Dec 9 '10 at 13:11

2 Answers 2

up vote 32 down vote accepted

Show that it's equal to $ { 2n \choose n } . $

Edit: Okay, here's how it goes:

\begin{align*} \left( 4 - \frac{2}{1} \right) \left( 4 - \frac{2}{2} \right) \left( 4 - \frac{2}{3} \right) \cdots \left( 4 - \frac{2}{n} \right) &=\small \frac{(4 \cdot 1 - 2)}{1} \frac{(4 \cdot 2 - 2)}{2} \frac{(4 \cdot 3 - 2)}{3} \cdots \frac{(4n - 2)}{n} \\ &= \small\frac{2^n (2 \cdot 1 - 1)(2 \cdot 2 - 1)( 2 \cdot 3 - 1) \cdots (2n-1)}{n!} \\ &= \frac{2^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \\ &= \frac{2^n \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdots (2n-1)(2n)}{n! \cdot 2 \cdot 4 \cdots (2n)} \\ &= \frac{2^n (2n)!}{ 2^n (n!)^2} = { 2n \choose n}. \end{align*}

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Ah, let me try. Thanks for the hint! –  anonymous Dec 9 '10 at 13:11
    
Nope. I am not getting anything. It would be helpful, if some more could be added. –  anonymous Jan 10 '11 at 20:47
    
Nice thanks a lot! –  anonymous Jan 11 '11 at 15:33
    
I think we can also use the fact that highest power of $2$ in $n!$ is less than $2^{n}$ to derive this result. –  user9413 May 22 '11 at 11:33

Using Pascal identity

$$\displaystyle\binom{2n}{n}=\left(4-\frac{2}{n}\right)\cdot\binom{2n-2}{n-1}$$

one gets for your product

$$\displaystyle\frac{\binom{2}{1}}{\binom{0}{0}}\cdot\frac{\binom{4}{2}}{\binom{2}{1}}\cdots\frac{\binom{2n-2}{n-1}}{\binom{2n-4}{n-2}}\cdot\frac{\binom{2n}{n}}{\binom{2n-2}{n-1}}=\binom{2n}{n}$$

and this is an integer as required.

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