# factorials

• Apr 4th 2011, 11:50 PM
salohcinseah
factorials
hi,

how do i do (2n-1)!/(2n+1)!
• Apr 5th 2011, 12:01 AM
veileen
$n!= 1 \cdot 2 \cdot 3\cdot...\cdot (n-1)\cdot n$
Write this way (2n-1)! and (2n+1)!, then simplify and see what remains.
• Apr 5th 2011, 12:11 AM
salohcinseah
so for (2n-1)! = 2n-1.2n where else for (2n+1)! = 2n+1.2n ?
• Apr 5th 2011, 12:18 AM
veileen
O.o No. You really read the first line?

n! = the product of first n natural unzero numbers. So... (2n-1)!=?
• Apr 5th 2011, 12:29 AM
amul28
salohcinseah,

I guess you din't get what veileen actually said.
See

$n! = 1. 2.\dots.(n-1).n$

So

$(2n-1)!= 1.2.\dots.(2n).(2n-1)$

Similarly

$(2n+1)!= 1.2.\dots.(2n-1).(2n).(2n+1)$

So, now divide both and see what remains.
• Apr 5th 2011, 12:31 AM
salohcinseah
i see so the answer will be 1/(2n+1)?
• Apr 5th 2011, 12:37 AM
veileen
amul28 - uhm, $2n > 2n-1$ >.>

$(2n-1)!=1 \cdot 2 \cdot...\cdot(2n-3)\cdot (2n-2)\cdot (2n-1)$

The answer will be $\frac{1}{2n(2n+1)}$.
• Apr 5th 2011, 12:40 AM
amul28