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Q.

1+xn=p0+p1x+p2x2++pnxn, then p0+p3+p6+=

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a

132n-1+cos3

b

132n-2+sin3

c

232n-2+sin6

d

232n-1+cos3

answer is B.

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Detailed Solution

(1+x)n=p0+p1x+p2x2+....+pnxn

Sub  x=1, w, w2

2n=p0+p1+p2+....+pn

(1+w)n=p0+p1w+p2w2+....

(1+w2)n=p0+p1w2+p2w+....

 2n+(1+w)n+(1+w2)n=3 (p0+p3+p6+....)

2n+1-12+i32n+1-12-i32n=3(p0+p3+p6+....)   

p0+p3+p6+...=132n+12+i32n+12-i32 =132n+cosπ3+i sinπ3n+cosπ3-i sinπ3n =132n+cos nπ3+i sin nπ3+cos nπ3-i sin nπ3 =132n+2cos nπ3

p0+p3+p6+==232n-1+cos3

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