Factorial identity question

  • Thread starter Paul245
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  • #1
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[itex]\frac{2 i}{(2 i + 1)!}[/itex] = [itex]\frac{1}{(2 i)!}[/itex] - [itex]\frac{1}{(2 i + 1)!}[/itex]

Could anybody please show what it is that needs to be done on LHS to get to RHS in this identity.
 

Answers and Replies

  • #2
coolul007
Gold Member
265
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[itex]\frac{2 i}{(2 i + 1)!}[/itex] = [itex]\frac{1}{(2 i)!}[/itex] - [itex]\frac{1}{(2 i + 1)!}[/itex]

Could anybody please show what it is that needs to be done on LHS to get to RHS in this identity.
(2i + 1)! = (2i+1)(2i!)
 
  • #3
13
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[itex]\frac{2i}{(2i+1)!}[/itex]=[itex]\frac{2i}{(2i+1)(2i)! }[/itex]
=[itex]\frac{2i}{2i(2i)! + (2i)! }[/itex]
= [itex]\frac{2i}{2i(2i)! + 2i (2i - 1)! }[/itex]
= [itex]\frac{1}{(2i)! + (2i - 1)! }[/itex]

and now what?
 
  • #4
45
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[itex]\frac{2i}{(2i+1)!}[/itex]=[itex]\frac{2i}{(2i+1)(2i)! }[/itex]
=[itex]\frac{2i}{2i(2i)! + (2i)! }[/itex]
= [itex]\frac{2i}{2i(2i)! + 2i (2i - 1)! }[/itex]
= [itex]\frac{1}{(2i)! + (2i - 1)! }[/itex]

and now what?
Nah, you've gone at that the wrong way.

Start with the question, and use coolul007's hint to get a common denominator on the right hand side.
 
  • #5
631
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Another way to go about: Add and subtract 1 in the numerator and club 2i and 1.
 
  • #6
13
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thanks acabus, all
 

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