If `(1+x+x^2)^n=a_0+a_1x+a_2x^2++a_(2n)x_(2n),` find the value of `a_0+a_6++ ,n in Ndot`

Explanation

`(1+x+x^(2)) = a_(0) + a_(1)x+a_(2)x^(2)+"……."+a_(2n)x^(2n)" "(1)`
In (1), putting `x = 1`, we get
`3^(n)=a_(0)+a_(1)+a_(2)+a_(3)+a_(4)+a_(5)+a_(6)+"….."" "(2)`
In (1), putting `x = omega` (imaginary cube root of unity), we get
`(1+omega+omega^(2))^(n)=a^(0)+a_(1)omega+a_(2)omega^(2)+a_(3)omega^(3)+a_(4)omega^(4)+a_(5)omega^(5)+a_(6)omega^(6)+"....."`
or `0=a_(0)+a_(1)omega+a_(2)omega^(2)+a_(3)+a_(4)omega+a_(5)omega^(2)+a_(6)+"....."" " (3)`
In (1), putting `x = omega^(2)`, we get
`(1+omega^(2)+omega^(4))^(n)=a_(0)+a_(1)omega^(2)+a_(2)omega^(4)+a_(3)omega^(6)+a_(4)omega^(8)+a_(5)omega^(10)+a_(6)omega^(12)+"......"`
or `0=a_(0)+a_(1)omega^(2)+a_(2)omega+a_(3)+a_(4)omega^(2)+a_(5)omega+a_(6)+"....."" "(4)`
Adding (2) and (3) and (4), we have
`3^(n) = 3(a_(0)+a_(3)+a_(6)+"....")`
or `a_(0)+a_(3)+a_(6)+"...."=3^(n-1)`