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,... ?
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
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?