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Prove that ^n C0+^n C3+^n C6+=1/3(2^n+2...

Prove that `^n C_0+^n C_3+^n C_6+=1/3(2^n+2cos(npi)/3)` .

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Consider `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)+.^(n)C_(2)+.^(n)C_(3)+.^(n)C_(4)+.^(n)C_(5)+.^(n)C_(6)+"....."`
`(1+omega)^(n) = .^(n)C_(0)+.^(n)C_(1)omega+.^(n)C_(2)omega^(2)+.^(n)C_(3)omega^(3)+.^(n)C_(4)omega^(4)+.^(n)C_(5)omega^(5)+.^(n)C_(6)omega^(6)+"......"`
`= .^(n)C_(0)+.^(n)C_(1)omega+.^(n)C_(2)omega^(2)+.^(n)C_(3)+.^(n)C_(4)omega+.^(n)C_(5)omega^(2)+.^(n)C_(6)+"......"`
`(1+omega^(3))^(n)= .^(n)C_(0)+.^(n)C_(1)omega^(2)+.^(n)C_(2)omega^(4)+.^(n)C_(3)omega^(6) +.^(n)C_(4)omega^(8)+.^(n)C_(5)omega^(10)+.^(n)C_(6)omega^(12)+"....."`
`= .^(n)C_(0)+.^(n)C_(1)omega^(2)+.^(n)C_(2)omega+.^(n)C_(3)+.^(n)C_(4)omega^(2)+.^(n)C_(5)omega + .^(n)C_(6)+"..."`
`:. 2^(n) +(1+omega)^(n)+(1+omega^(2))^(n) = 3(.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+".....")`
Now, `2^(n)+(1+omega)^(n)+(1+omega^(2))^(n)=2^(n)+2Re((1+omega)^(n))`
`= 2^(n) +2Re(1/2-i'(sqrt(3))/(2))^(n)`
`= 2^(n)+2Re(cos'(pi)/(3)-isin'(pi)/(3))^(n)`
`= 2^(n)+2cos'(npi)/(3)`
Hence, `.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+"...."=1/3(2^(n)+2cos'(npi)/(3))`
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