# sequences 1

• Aug 30th 2010, 02:41 AM
furor celtica
sequences 1
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
• Aug 30th 2010, 03:15 AM
Moo
Hello,

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
• Aug 30th 2010, 04:03 AM
furor celtica
um, this is not so clear, what do you mean by
'you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1'
?
• Aug 30th 2010, 08:11 AM
Soroban
Hello, furor celtica!

Quote:

$\text{Show that: }\;\dfrac{(2n)!}{n!} \:=\: 2^n(1\cdot3\cdot 5 \cdots [2n -1])$

Did you examine a specific case?

Let $n = 4.$

Then we have: . $\dfrac{8!}{4!} \;=\;\dfrac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\ cdot1}{4\cdot3\cdot2\cdot1}$

Reduce: . $\dfrac{\not8^{^2}\! \cdot 7 \;\cdot\!\not6^{^2}\! \cdot5\; \cdot \!\not4^{^2}\! \cdot3 \;\cdot \!\not2^{^2}\cdot 1}{\not4\; \cdot\! \not3 \;\cdot\! \not2 \;\cdot\! \not1}$

. . and we have: . $2^4(1\cdot3\cdot5\cdot7)$

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?

• Aug 30th 2010, 08:37 AM
CaptainBlack
Quote:

Originally Posted by furor celtica
um, this is not so clear, what do you mean by
'you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1'
?

You separate the $(2n)!$ factorial into the product of odd and even terms

CB
• Sep 1st 2010, 11:19 PM
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?