# Double Factorial

1. Oct 14, 2011

### matematikuvol

How to prove

$$(2n)!!=2^nn!$$

for example

$$(2)!!=2^11!=2$$

$$(4)!!=2^22!=8$$

...

$$(2n)!!=2n(2n-2)(2n-4)...=2^3n(n-1)(n-2)...$$

I see that by intiution but I don't know how to write prove.

2. Oct 14, 2011

### Mute

Prove it by induction. You've shown that it holds for the case n =1, so if you assume that it holds for general n, show that it follows that it holds for the case n+1.

3. Oct 14, 2011

### dextercioby

You don't need induction. A direct computation using the definition of the double factorial is enough.

4. Oct 14, 2011

### matematikuvol

How?

5. Oct 14, 2011

### Dickfore

$$(2 n)!! = 2 \times 4 \times \ldots \times (2 n)$$

But, notice that:
$$\begin{array}{l} 2 = 2 \times 1 \\ 4 = 2 \times 2 \\ \ldots \\ 2 n = 2 \times n \end{array}$$

Combine the first factor of two from each factor in the double factoriel. How many of them are there? What do the remaining factors give?

6. Oct 14, 2011

Tnx :)