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Question

If (1+x+x2)n=a0+a1x+a2x+a3x3+.......a(2n)x(2n) then prove that(a0+a3+a6+.........)=(a1+a4+a7)=(a2+a5+a8) =3(n1),

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Solution

put =ωandnotethat1+ω+ω2=0
where ω=12+i32andω2=12i32
Δ=(a0+a3+a6+......)+ω(a1+a4+......)+ω(a2+a3......)
or Δ=Aω+Bω2+C=0
equating real and imaginary part of both sides
AB2C2=032(BC)=0
B = C and hence A = B
A \, = \, B \, = \., C
Again putting x = 1 in the given relation . we get 3^ n = sum of the all the condition = A + B + C

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