If `(1+x+x^(2))^(3n+1)=a_(0)+a_(1)x+a_(2)x^(2)+…a_(6n+2)x^(6n+2)`, then find the value of `sum_(r=0)^(2n)(a_(3r)-(a_(3r+1)+a_(3r+2))/2)` is

Answer

0

Explanation

`(1+x+x^(2))^(3n+1)=a_(0)+a_(1)x+a_(2)x^(2)+….a_(6n)+a_(6n+1)omega+a_(6n+2)omega^(2)`………1
`(1+omega^(2)+omega^(4)=0=a_(0)+a_(1)omega^(2)+a_(2)omega+a_(3)+…..+a_(6n+1)omega^(2)+a_(6n+2)omega`….2
Adding 1 and 2 we get
`sum_(r=0)^(2n)(2a_(3r)-(a_(3r+1)+a_(3r+2)))=0`