Qualification > Math

Math again urrrgghh

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astarmathsandphysics:
I will write some as soon as i know for what module.

nid404:

--- Quote from: astarmathsandphysics on September 20, 2009, 03:33:10 pm ---I will write some as soon as i know for what module.

--- End quote ---

Thanks :)

Ghost Of Highbury:
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
.......

nid404:

--- 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
.......




--- End quote ---

Very confusing adi

Thanx for your help anyways :)

Ghost Of Highbury:

--- Quote from: nid404 on September 21, 2009, 04:41:55 am ---Very confusing adi

Thanx for your help anyways :)

--- End quote ---

i found this much easier thank k+1

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