Double Factorial

  • #1
How to prove

[tex](2n)!!=2^nn![/tex]

for example

[tex](2)!!=2^11!=2[/tex]

[tex](4)!!=2^22!=8[/tex]

...

[tex](2n)!!=2n(2n-2)(2n-4)...=2^3n(n-1)(n-2)...[/tex]

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

Answers and Replies

  • #2
Mute
Homework Helper
1,388
10
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
dextercioby
Science Advisor
Homework Helper
Insights Author
13,002
552
You don't need induction. A direct computation using the definition of the double factorial is enough.
 
  • #5
2,967
5
[tex]
(2 n)!! = 2 \times 4 \times \ldots \times (2 n)
[/tex]

But, notice that:
[tex]
\begin{array}{l}
2 = 2 \times 1 \\

4 = 2 \times 2 \\

\ldots \\

2 n = 2 \times n
\end{array}
[/tex]

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
Tnx :)
 

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