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posted by  justbecaz on 8/26/2008 11:18:29 AM  |  status: Live  

Number Theory - PMI Lifesaving Rating URGENT URGENT URGENT

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Verify that for all n ≥ 1,

 

2*6*10*14… (4n-2) = (2n)! /n!

 

Basic Step:

(4(1)-2) = (2(1)1!/1!

          2 = 2

 

Assume P(k) is true

Show that P(k+1) is true

2*6*10*14… (4n-2)*(4(n+1)-2) = (2n+1)! /(n+1)!

 

Consider:

2*6*10*14 … (4n-2) * (4n+2) = (2n+1)! /(n+1)!

 

Okay this is where my confusion comes in, if I am correct up to this point.

I know that on the left side 2*6*10*14 … (4n-2) = (2n)!/n!

So now is my equation : (2n)!/n! *(4n+2) =(2n+1)! /(n+1)!

 

Ok have I proceeded correctly so far and if not where am I going wrong, what should my next step be.

 

Any help would be greatly appreciated.

Tags: Math, Other
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posted by zsm28 on 8/26/2008 1:44:06 PM  |  status: Live
Asker's Rating: Lifesaver   
justbecaz's comment:
"Thank you so much, I knew I had done something wrong and that was it thanks a million"
Response Details:
In your post

Assume P(k) is true

Show that P(k+1) is true

2*6*10*14… (4n-2)*(4(n+1)-2) = (2n+1)! /(n+1)!

BUT IT IS WRONG!!!

it should be:

Show that P(k+1) is true

2*6*10*14… (4n-2)*(4(n+1)-2) = (2(n+1))! /(n+1)!

so you need to prove:
(2n)!/n! *(4n+2) =(2n+2)! /(n+1)!
It's easy:
(2n)!/n! *(4n+2) = 2*(2n)!*(2n+1)/n! = 2*(2n+1)!*(n+1)/(n+1)! = (2n+1)!*(2n+2)/(n+1)! = (2n+2)!/(n+1)!

Tags: Math, Other
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