hey i'm truly sorry if this isnt in the right forum, i actually couldnt find a subforum that addressed sequence problems
so this is a basic problem inolving factorials so don't make the answer too complicated please
show that (2n)!/n! = 2^n (1x3x5...x(2n -1))
i really have very little work to show cos i got nowhere in the hours i've been trying to wrap my head around this. its probably simple and i'm overestimating it but the farthest i've got is to 2n x (2n-1) x (2n-2) x (2n-3) x...x (n)
i'm not even sure that is correct
any help would be appreciated, please show all working and details, i'd like to totally understand this
In (2n)! you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1
Now what if you factor out 2 of each of 2n,2n-2,... ? :D
um, this is not so clear, what do you mean by
'you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1'
Hello, furor celtica!
Did you examine a specific case?
Then we have: .
. . and we have: .
Do you see what happened?
The denominator cancels into the even factors in the numerator
. . leaving behind a "bunch of 2's" ... and all the odd factors.
Now can you express all that in general terms?
You separate the factorial into the product of odd and even terms
Originally Posted by furor celtica
ok so i started with the RHS and got to ((2^n)(2n!))/(2x4x6x...x2n), i.e. for the statement to be true i have to prove that (2x4x6x...x2n) = (2^n)(n!)
does this need to be proven really or is it obvious?