Question

can anybody demostrate this please ? ((2n)!)/(n!)= 2^n(1x3x5x7.....(2n-1))

Answer 1

http://www.mae.ufl.edu/~uhk/factorial.pdf

Answer 2

tried to "unzip"...
2n*(2n+1)*(2n+2)(2n+3)(2n+4)(2n+5)(2n+6)...
you can re-write it as:
2n(2n+1)*2(n+1)(2n+3)*2*(n+2)(2n+5)*2*(n+3)(....

just looking at the sequence you can see that n will can canceled out
will canceled out all even multipliers (un-zip your denominator: n*(n+1)(n+2)(n+3)...)

all that's left:
2^n * (odd numbers)

would it work?

Answer 3

i think it may work ... thanks :D

Answer 4

its n(n-1)(n-2)...., not n(n+1)(n+2)....

Answer 5

i think it depends what you used to. i do not see any difference how you'll write it...
for example:
5!=1*2*3*4*5 or = 5*4*3*2*1
the result is the same... right?
I used (only because it easier to calculate & canceled out):
n!=n*(n+1)(n+2)(n+3)....
n>=0
so, (2n)!=2n*(2n+1)(2n+2)(2n+3)(2n+4)....

Answer 6

no. that is incorrect.
if you say n! = n*(n+1)(n+2)(n+3)....
then 5! would be 5*(5+1)(5+2)(5+3)...
that is, in your case, 5! = 5*6*7*8..... where does it stop? at infinity?

Answer 7

no...
I wrote that n>=0
"my" n starts from 0 & going up...to infinity in this case

Answer 8

if you take n(n-1)(n-2)...
and (2n)(2n-1)*2n-2)(2n-3)....
the result will be the same
the denominator will cancel out, 2 will get ^n and every all number (2n-1)(2n-3)(2n-5)... will be left
right?

Answer 9

n!=n*(n+1)(n+2)(n+3)....
this was your statement. correct?

Answer 10

that is an incorrect statement. it does not make sense as a definition of factorial and would work very little of the time.

Answer 11

i see what you saying...
agreed :)
do you agree that
the denominator will cancel out, 2 will get ^n and every all number (2n-1)(2n-3)(2n-5)... will be left in your case?

Answer 12

okk.... now that u two agree in that, could you tell me the right answer? xD

Answer 13

your definition, as I said will work very little of the time and is not a definition of factorial at all

Answer 14

it's "almost" the same... :) (change sign)
your present numerator as:
2n*(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)....
2n*(2n-1)*2(n-1)(2n-3)*2*(n-2)(2n-5)*2*(n-3)...

and denominator as:
n(n-1)(n-2)(n-3)(n-4)....

as you can see n, (n-1), (n-2) ... are canceling out
what is left in numerator :
2^n *(2n-1)(2n-3)(2n-5)*.... (odd numbers)

isn't it what you want?

Answer 15

yess thanks :D

Answer 16

wait if i cancel n in the denominator with the 2n in the numerator it stay like this:
2*(2n-1)(2n-3)....

Answer 17

don't forget 2 *2*2*2... from (common multiplier):
2n*(2n-1)*2(n-1)(2n-3)*2*(n-2)(2n-5)*2*(n-3)...

Answer 18

ok... but i need the n out of the parentheses

Answer 19

what do you mean?

Answer 20

2n(2n-1)*2(n-1)(2n-3)*2*(n-2)(2n-5)*2*(n-3)(2n-7)...
------------------------------------------------
n (n-1) (n-2) (n-3)...

do you see it?
n/n , (n-1)/(n-1), (n-2)/(n-2)....
and all that's left:
2*2*2*2*...(2n-1)(2n-3)(2n-5)*...

Answer 21

yes 2*2*2*2..... =2^n
but i need (1x3x5x7.... i dont see it :S

Answer 22

that is:
(2n-1)(2n-3)(2n-5)...
it's odd numbers...
for example:
if n=20
39*37*35*33....*5*3*1

it just in opposite order that in your problem

Answer 23

oww thanks :D

Answer 24

welcome!
this only "simple" way that I can see it ... without bringing "big guns" & formulas :)