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Sir pls solve the following.
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Asked by rsudipto | 29 Dec, 2018, 11:47: AM
answered-by-expert Expert Answer
Given expression :  (1+i)n1 + (1+i3)n1 + (1+i5)n2 + (1+i7)n2 .......................(1)
 
since i = √-1  , i3 = - i  , i5 = i  and i7 = - i
 
Hence eqn.(1) becomes,  (1+i)n1 + (1-i)n1 + (1+i)n2 + (1-i)n2
 
begin mathsize 12px style f o r space a n y space p o s i t i v e space i n t e g e r space n comma space u sin g space b i n o m i a l space e x p a n s i o n w e space h a v e space left parenthesis 1 plus i right parenthesis to the power of n space equals space 1 space plus space C presuperscript n subscript 1 space i space minus space C presuperscript n subscript 2 space minus C presuperscript n subscript 3 space i space plus space C presuperscript n subscript 4 space plus space C presuperscript n subscript 5 space i space minus C presuperscript n subscript 6 space........ w e space h a v e space left parenthesis 1 minus i right parenthesis to the power of n space equals space 1 space minus space C presuperscript n subscript 1 space i space minus space C presuperscript n subscript 2 space plus C presuperscript n subscript 3 space i space plus space C presuperscript n subscript 4 space minus space C presuperscript n subscript 5 space i space minus C presuperscript n subscript 6 space........ H e n c e space left parenthesis 1 plus i right parenthesis to the power of n space plus space left parenthesis 1 minus i right parenthesis to the power of n space i s space r e a l space f o r space a n y space n end style
hence in the option list, (D) is correct answer,   i.e., n1>0 , n2 >0
Answered by Thiyagarajan K | 30 Dec, 2018, 04:12: PM

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