Proving Double Factorial: (2n)!=2^nn!

[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.

 

[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.

 

[dextercioby]

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

 

[matematikuvol]

How?

 

[Dickfore]

<br /> (2 n)! = 2 \times 4 \times \ldots \times (2 n)<br />

But, notice that:
<br /> \begin{array}{l}<br /> 2 = 2 \times 1 \\<br /> <br /> 4 = 2 \times 2 \\<br /> <br /> \ldots \\<br /> <br /> 2 n = 2 \times n<br /> \end{array}<br />

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?

 

[matematikuvol]

Tnx :)