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[Post New]2 Apr 2008 10:00:45 IST
    Subject: Re:good question
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ramkumar_november (1270)

Blazing goIITian

Olaaa!! Perrrfect answer. 230  [290 rates]
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here is one method that i have found out......
 
LET
 
C;=;cosx+nC_1cos2x+nC_2cos3x+.........+nC_ncos(n+1)x
 
S;=;sinx+nC_1sin2x+nC_2sin3x+.........+nC_nsin(n+1)x
 
now
 
C+iS;=;(cosx+isinx)+nC_1(cos2x+isin2x)+....+cos(n+1)x+isin(n+1)x
 
using euler's formula...  e^{ix};=;cosx+isinx
 
C+iS;=;e^{ix}+nC_1e^{i2x}+nC_2e^{i3x}+....+e^{i(n+1)x}
 
C+iS;=;e^{ix}(1+nC_1e^{ix}+nC_2e^{i2x}+....+e^{inx})
 
C+iS\;=\;e^{ix}(1+e^{ix})^n
 
see hsbhatt sir's post after this........
 
 the answer is [C].......
 
 
 this reply: 20 points  (with 4 Olaaa!! Perrrfect answer.   in 4 votes )   [?]
 
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