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