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^(2)`
`(3+omega+omega^(2))^(2010) = a_(0) + a_(1)omega + a_(2)omega^(2)+"….."`
`rArr 2^(210) = a_(0) +a_(1)omega + a_(2)omega^(2) + a_(3) + a_(4)omega + "……"`
and
`2^(210) = a_(0) + a_(1)omega^(2) + a_(2)omega^(2) + a_(2)omega+a_(3) + a_(4)omega^(2) + "....."`
Adding (1) and (2), we have
`2 xx 2 ^(210) = 2a_(0) - a_(1) - a_(2) + 2a_(3) - a_(4) - a_(5) + 2_(6^(-))`
`rArr 2^(210) = a_(0) - 1/(2) a_(1) - 1/(2) a_(2) + a_(3) - 1/(2) a_(4) - 1/(2) a_(5) + a_(6) "...."`