IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: nid404 on September 20, 2009, 01:12:12 pm

Title: Math again urrrgghh
Post by: nid404 on September 20, 2009, 01:12:12 pm: Show that (2n)!/n!=2ⁿ (1x3x5x....x(2n-1))

I don't even understand the answer....what's the stuff in the brackets supposed to mean?
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 20, 2009, 01:18:08 pm: baap re!

will try..:P
Title: Re: Math again urrrgghh
Post by: nid404 on September 20, 2009, 02:05:16 pm: Quote from: eddie_adi619 on September 20, 2009, 01:52:42 pm
okk..got it...

first simplify this part..which u didnt understand..:P

(1x3x5x....x(2n-1))

1*3*5... = 1*2*3*4*5*6........
------------------- ------------------ = n! / n!! = 1/n!
~~2*4*6~~*8......

R.H.S ==> 2ⁿ / n!

= (2n)! / n!

didn't get it :(....what is n!!...n factorial factorial.....no no im confused nw
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 20, 2009, 02:10:42 pm: nono..sry..that was incorrect...will correct it

i read it in a pdf...and wiki....
Title: Re: Math again urrrgghh
Post by: nid404 on September 20, 2009, 02:33:46 pm: To show or rather prove this, can I use a value for n...is it ok

If I do using a value i get

it's like

for example I use n=6

then 12!/6!=2^6(1x3x5x7x9x11)which is all the odd no from 1 to (2n-11)
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 20, 2009, 02:36:20 pm: Quote from: nid404 on September 20, 2009, 02:33:46 pm
To show or rather prove this, can I use a value for n...is it ok

If I do using a value i get

it's like

for example I use n=6

then 12!/6!=2^6(1x3x5x7x9x11)which is all the odd no from 1 to (2n-11)

umm..nono...u cant do that....u have to simplify the LHS to get the RHS or vice versa...i'm still trying...
Title: Re: Math again urrrgghh
Post by: nid404 on September 20, 2009, 02:43:29 pm: gosh i need astar to do this
Title: Re: Math again urrrgghh
Post by: astarmathsandphysics on September 20, 2009, 02:44:58 pm: I will answer this now. Have been writing maths notes all day.
Title: Re: Math again urrrgghh
Post by: astarmathsandphysics on September 20, 2009, 02:59:42 pm: Quote from: nid404 on September 20, 2009, 01:12:12 pm
Show that (2n)!/n!=2ⁿ (1x3x5x....x(2n-1))

I don't even understand the answer....what's the stuff in the brackets supposed to mean?
5!=5*4*3*2*1

Prover by induction
If n=1, $\frac {(2n)!}{n!}={2^1}*(2*1-1)$ so true for n=1. Suppose true for n=k. Prover true for n=k+1
$\frac {(2(k+1))!}{(k+1)!}=\frac {(2k+2)(2k+1)(2k)!}{(k+1)k!}=\frac {2(k+1)(2k+1)(2k)!}{(k+1)k!}=\frac {2(2k+1)(2k)!}{k!}=2(2k+1) \frac {(2k)!}{k!}$

Now we use that (2k)!/k!=2^k (1x3x5x....x(2k-1))

$2(2k+1) \frac{(2k)!}{k!}=2(2k+1)2^k (1x3x5x....x(2k-1))=2^{k+1}(1x3x5x....x(2k-1)(2k+1))$
Title: Re: Math again urrrgghh
Post by: nid404 on September 20, 2009, 03:08:22 pm: Thank you very much...I kinda got it but I need practice and some notes too
Title: Re: Math again urrrgghh
Post by: astarmathsandphysics on September 20, 2009, 03:33:10 pm: I will write some as soon as i know for what module.
Title: Re: Math again urrrgghh
Post by: nid404 on September 20, 2009, 03:33:46 pm: Quote from: astarmathsandphysics on September 20, 2009, 03:33:10 pm
I will write some as soon as i know for what module.

Thanks :)
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 21, 2009, 03:57:45 am: got this from yahoo answers....
L.H.S. = (2n)! / n!

=> [1× 2 × 3 × 4 × 5 × . . . . . . . . (2n - 2) (2n - 1) (2n) ]/ n!

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × [ 2 × 4 × 6 × 8 . . . (2n - 2) (2n) ] / n!
{rearranging odd & even terms }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × 2^n [ 1× 2 × 3 × 4 × 5 . . .(n - 1) n ]/n!
{taking 2, n times common from 2nd bracket }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × [2^n × n! ] / n!
{As ,1× 2 × 3 × 4 × 5 . . .(n - 1) n = n! }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × 2^n
.......
Title: Re: Math again urrrgghh
Post by: nid404 on September 21, 2009, 04:41:55 am: Quote from: eddie_adi619 on September 21, 2009, 03:57:45 am
got this from yahoo answers....
L.H.S. = (2n)! / n!

=> [1× 2 × 3 × 4 × 5 × . . . . . . . . (2n - 2) (2n - 1) (2n) ]/ n!

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × [ 2 × 4 × 6 × 8 . . . (2n - 2) (2n) ] / n!
{rearranging odd & even terms }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × 2^n [ 1× 2 × 3 × 4 × 5 . . .(n - 1) n ]/n!
{taking 2, n times common from 2nd bracket }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × [2^n × n! ] / n!
{As ,1× 2 × 3 × 4 × 5 . . .(n - 1) n = n! }

=> [1 × 3 × 5 × 7 . . . (2n - 1) ] × 2^n
.......

Very confusing adi

Thanx for your help anyways :)
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 21, 2009, 04:44:00 am: Quote from: nid404 on September 21, 2009, 04:41:55 am
Very confusing adi

Thanx for your help anyways :)

i found this much easier thank k+1
Title: Re: Math again urrrgghh
Post by: nid404 on September 21, 2009, 04:46:45 am: Quote from: eddie_adi619 on September 21, 2009, 04:44:00 am
i found this much easier thank k+1

but this is not linked correctly
why would you simply do it without a reason
as in

[1× 2 × 3 × 4 × 5 × . . . . . . . . (2n - 2) (2n - 1) (2n) ]/ n!

how does 2n!/n! become something like this....no...not getting it
Title: Re: Math again urrrgghh
Post by: Ghost Of Highbury on September 21, 2009, 04:49:50 am: arre dude...

n! = 1 * 2 * 3 * 4 ......*n rite?

so (2n)! = 1*2*3*4.......2n..

suppose n = 3

n! = 1*2*3 = 6

2n = 6

6! = (2n)! = 1*2*3*4*5*6. = 720..

thats the rule///
Title: Re: Math again urrrgghh
Post by: nid404 on September 21, 2009, 04:54:21 am: I know that.....

but still note getting what is done there