hi,

how do i do (2n-1)!/(2n+1)!

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- Apr 4th 2011, 11:50 PMsalohcinseahfactorials
hi,

how do i do (2n-1)!/(2n+1)! - Apr 5th 2011, 12:01 AMveileen
$\displaystyle n!= 1 \cdot 2 \cdot 3\cdot...\cdot (n-1)\cdot n$

Write this way (2n-1)! and (2n+1)!, then simplify and see what remains. - Apr 5th 2011, 12:11 AMsalohcinseah
so for (2n-1)! = 2n-1.2n where else for (2n+1)! = 2n+1.2n ?

- Apr 5th 2011, 12:18 AMveileen
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 AMamul28
salohcinseah,

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

See

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

So

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

Similarly

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

So, now divide both and see what remains. - Apr 5th 2011, 12:31 AMsalohcinseah
i see so the answer will be 1/(2n+1)?

- Apr 5th 2011, 12:37 AMveileen
amul28 - uhm, $\displaystyle 2n > 2n-1$ >.>

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

The answer will be $\displaystyle \frac{1}{2n(2n+1)}$. - Apr 5th 2011, 12:40 AMamul28
my bad.

you are correct.

salohcinseah, please correct that.