Factorial identity question

  • Thread starter Paul245
  • Start date
[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.
 

coolul007

Gold Member
262
7
[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!)
 
[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?
 
45
0
[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.
 
Another way to go about: Add and subtract 1 in the numerator and club 2i and 1.
 
thanks acabus, all
 

Related Threads for: Factorial identity question

  • Posted
Replies
2
Views
2K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
2
Views
396
  • Posted
Replies
3
Views
3K
Replies
1
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top