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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
10 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
10 years ago

Dear Shashank,

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

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

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shashank
38 Points
10 years ago

this is not factorrial ans is 2 explain how

 

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