If `(3+x^(2008)+x^(2009))^(2010)=a_0+a_1x+a_2x^2++a_n x^n ,` then the value of `a_0-1/2a_1-1/2a_2+a_3-1/2a_4-1/2a_5+a_6-` is `3^(2010)` b. `1` c. `2^(2010)` d. none of these

Options

A. `3^(2010)`
B. 1
C. `2^(2010)`
D. none of these

Answer

C

Explanation

Put `x = omega, omega`
`(3+omega+omega^(2))^(2010) = a_(0)+a_(1)omega+a_(2)omega^(2)+"....."`
and `(3+omega^(2)+omega^(4))=a_(0)+a_(1)omega^(2)+a_(2)omega^(4)+"...."`
`2^(2010) = a_(0)+a_(1)omega+a_(2)omega^(2)+a_(3)+a_(4)omega+"....."" "(1)`
and `2^(2010) = a_(0) + a_(1)omega^(2)+a_(2)omega+a_(3)+a_(4)omega^(2)+"......"(2)`
Adding (1) and (2), we have
`2 xx 2^(2010) = 2a_(0) - a_(1) - a_(2) + 2a_(3)-a_(4)-a_(5) + 2a_(6) - "......"`
or `2^(2010) = a_(0) - (1)/(2)a_(1) - 1/2a_(2) + a_(3) - 1/2 a_(4) - 1/2a_(5) + a_(6) "....."`