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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: PLEAZ HELP..............
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priyeshpatel (0)

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SHOW THAT
(1-  + 2) ( 1- 2 + 4) (1- 48 )..........TO  2N FACTORS  = 2N
    
sriram4frendz (49)

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1+w+w^2=0
1+w^2=-w
sub in eqn similarly w^4=w therefore 1+w=-w^2,w^8=w^2therfroe(1-w^4+w^8)=(-2w)
(-2w)(-2w^2)(-2w)...........upto2n factors=2^n
hence proved(becuase first two terms becomes 4 hence all terms will become four
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Greatdreams (3307)

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(1 - w + w^2)(1- w^2 + w^4)(1 - w^4 + w^8)............to 2n factors

= ( 1 - w + w^2)(1 - w^2 + w)(1 - w + w^2)..........upto 2n factors

( since, w^4 = w , w^8 = w^2 , w^16 = w.......)

=(-2w)(-2w^2)(-2w)(-2w^2)............to 2n factors

= (2^2w^3)(2^2w^3)(2^2w^3).........upto 2n factors

= 2 ^2n.......proved

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chimanshu_007 (11609)

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yes the ans comes 22n

see the above method

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