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Q.

nC0+nC3+⋯−(1/2) nC1+nC2+nC4+nC5+⋯2+(3/4) nC1−nC2+nC4−nC5+⋯2=

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a

3

b

4

c

2

d

1

answer is D.

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Detailed Solution

(1+ω)n=nC0+nC1ω+⋯ = nC0+nC3+⋯+ nC1+nC4+⋯−1+3i2+ nC2+nC5+⋯−1−3i2 = nC0+nC3+⋯−12 nC1+nC2+nC4+nC5…+i32 nC1−nC2+nC4−nC5+⋯Equating the modulus, we get −ω2n=1. ( If z = x +iy then z2=x2+y2 )

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nC0+nC3+⋯−(1/2) nC1+nC2+nC4+nC5+⋯2+(3/4) nC1−nC2+nC4−nC5+⋯2=