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Question

lf (1+x+x2)n=a0+a1x+a2x2++a2nx2n, then a0+a3+a6+...=

A
3n
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B
3n1
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C
3n2
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D
1
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Solution

The correct option is A 3n1
Substituting x=1 we get,
3n=a0+a1+a2+a3...+a2n...(i)
Substituting x=ω we get, (where w is the non-real cube root of unity)
(1+ω+ω2)=0
a0+a1(ω)+a2(ω2)+a3(ω3)+...+a2n(ω2n)=0...(ii)
Substituting x=ω2 we get,
(1+ω2+ω4)
=(1+ω2+ωω3)
=(1+ω2+ω)
=0
a0+a1ω2+a2ω4+a3ω6...a2nω4n=0...(iii)
Adding (i), (ii) and (iii) we get
3n=3a0+a1(1+ω+ω2)+a2(1+ω2+ω4)+a3(1+ω3+ω6)...
3n=3a0+a2(1+ω2+ω3ω)+a3(1+1+ω3ω3)...
3n=3(a0+a3+...)
a0+a3+a6...=3n1

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