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the smallest +ve number nfor which (1-i)^2n=(1+i)^2n is

the smallest +ve number nfor which (1-i)^2n=(1+i)^2n is

Grade:12

3 Answers

Aman Bansal
592 Points
9 years ago

Dear Shashank,

first compare (2n-1)! and (2n+1) and forget the division for right now

the first 5 terms of (2n-1) are --> (2n-1)(2n-2)(2n-3)(2n-4)(2n-5)...
the first 5 terms of (2n+1) are --> (2n+1)(2n)(2n-1)(2n-2)(2n-3)...
so you can see that after the first two terms of (2n+1)!, it is exactly identical to (2n-1)!, so therefore everything except (2n+1) and (2n) cancel out, resulting in the answer 1 / 2n(2n+1)

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Aman Bansal
592 Points
9 years ago

Dear Shashank,

Please dont post same query repeatedly.....!!!!

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Aman Bansal

Askiitian Expert

 

shashank
38 Points
9 years ago

this is not factorrial ans is 2 explain how

 

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