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If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},$ then value of $a_{0} + a_{3} + a_{6} + \dots$ is

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a

1

b

2n

c

2n-1

d

3n-1

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detailed solution

Correct option is D

Putting x=1,ω,ω2 in 1+x+x2n=a0+a1x+a2x2+…+a2nx2n, we get (1+1+1)n=a0+a1+a2+a3+…+a2n,1+ω+ω2n=a0+a1ω+a2ω2+a3ω3+…+a2nω2n,and 1+ω2+ω4n=a0+a1ω2+a2ω4+a3ω6+…+a2nω4nAdding the above three equations and using 1+ω+ω2=0, ω3=1 we get 3n=3(a0+a3+a6+....)⇒ a0+a3+a6+...=3n-1

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