- #1

- 192

- 0

[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.

- Thread starter matematikuvol
- Start date

- #1

- 192

- 0

[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.

- #2

Mute

Homework Helper

- 1,388

- 10

- #3

- 13,002

- 552

- #4

- 192

- 0

How?

- #5

- 2,967

- 5

(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

- 192

- 0

Tnx :)

- Last Post

- Replies
- 1

- Views
- 13K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 481

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 6K

- Last Post

- Replies
- 12

- Views
- 843

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 3K