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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 |
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Answer» `(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` |
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