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Author Topic: Math again urrrgghh  (Read 2883 times)

nid404

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Math again urrrgghh
« 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?
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Offline Ghost Of Highbury

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Re: Math again urrrgghh
« Reply #1 on: September 20, 2009, 01:18:08 pm »
baap re!

will try..:P
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nid404

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Re: Math again urrrgghh
« Reply #2 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
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Offline Ghost Of Highbury

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Re: Math again urrrgghh
« Reply #3 on: September 20, 2009, 02:10:42 pm »
nono..sry..that was incorrect...will correct it




i read it in a pdf...and wiki....

« Last Edit: September 20, 2009, 02:17:27 pm by eddie_adi619 »
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nid404

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Re: Math again urrrgghh
« Reply #4 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)


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Offline Ghost Of Highbury

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Re: Math again urrrgghh
« Reply #5 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...
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nid404

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Re: Math again urrrgghh
« Reply #6 on: September 20, 2009, 02:43:29 pm »
gosh i need astar to do this
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Offline astarmathsandphysics

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Re: Math again urrrgghh
« Reply #7 on: September 20, 2009, 02:44:58 pm »
I will answer this now. Have been writing maths notes all day.
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Offline astarmathsandphysics

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Re: Math again urrrgghh
« Reply #8 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!=2k (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))
« Last Edit: September 20, 2009, 03:06:57 pm by astarmathsandphysics »
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nid404

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Re: Math again urrrgghh
« Reply #9 on: September 20, 2009, 03:08:22 pm »
Thank you very much...I kinda got it but I need practice and some notes too
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Offline astarmathsandphysics

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Re: Math again urrrgghh
« Reply #10 on: September 20, 2009, 03:33:10 pm »
I will write some as soon as i know for what module.
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nid404

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Re: Math again urrrgghh
« Reply #11 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 :)
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Offline Ghost Of Highbury

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Re: Math again urrrgghh
« Reply #12 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
.......


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nid404

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Re: Math again urrrgghh
« Reply #13 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 :)
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Offline Ghost Of Highbury

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Re: Math again urrrgghh
« Reply #14 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
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