sequences 1

[red and white kop!]

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

 

[Deleted member 4993]

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

Answer cannot be made too complicated - this is basic arithmatic

Multiply by 1

\(\displaystyle 2^n [1*3*5...*(2n -1)] \ = \ 2^n [1*3*5...*(2n -1)] \ \ * \ \frac {2*4*6*....*(2n)}{2*4*6*....*(2n)}\)

now continue.... (and quit using 'x' for multiplication - use '*' instead)

 

[red and white kop!]

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?

 

[BigGlenntheHeavy]

\(\displaystyle Prove \ \frac{(2n)!}{n!} \ = \ 2^n[1*3*5*...*(2n-1)]\)

\(\displaystyle Now, \ following \ Subhotosh \ Khan's \ lead, \ we \ get: \ 2^n[1*3*5*...*(2n-1)] \ * \ \frac{2*4*6*...*(2n)}{2*4*6*...*(2n)}\)

\(\displaystyle This \ equals \ \frac{2^n[1*2*3*4*5*6*...*(2n-1)(2n)]}{2*4*6*...*(2n)} \ = \ \frac{2^n[1*2*3*4*5*6*...*(2n-1)(2n)]}{2^n[1*2*3*4*5*6*...*n]}\)

\(\displaystyle = \ \frac{(2n)!}{n!}, \ QED\)