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

Options

A. 1
B. `2^(2010)`
C. `5^(2010)`
D. `3^(2010)`

Answer

B

Explanation

Given ,`(3+ x^(2008) + x^(2009) + x^(2009))^(2010) = a_(0) + a_(1) x + a_(2)x^(2) + ...+a_(n)x^(n)`
On putting ` x = omega " and " omega^(2)` respectively, we get
`(3+ x^(2008) + x^(2009) + x^(2009))^(2010) = a_(0)omega+ a_(2)omega^(2) + a_(3) omega^(3) + ...`
or ` (3+ omega^(2008)+ omega^(2))^(2010) = a_(0) + a_(1) omega + a_(2) omega^(2) + a_(3)omega^(3) + ...`
or ` 2^(2010) a_(0) + a_(1) omega + a_(2) omega^(2) + a_(3) omega^(3) + a_(4)omega^(4) + a_(5)omega^(5) + a_(6) omega^(6) + ...` (i)
and `[+ (omega^(2))^(2008) + (omega^(2))^(2009)]^(2010)`
` = a_(0) + a_(1) omega^(2) + a_(2) omega^(4) + a_(3) omega^(6) + a_(4) omega^(8) +a_(5)omega^(10) + a_(6) omega^(12) + ...`
or ` (3+ omega^(2) + omega)^(2010)`
` rArr = a_(0) + a_(1) omega^(2) + a_(2) omega^(4) + a_(3) omega^(6) + a_(4) omega^(8) +a_(5)omega^(10) + a_(6) omega^(12) + ...` (ii)
On adding Eqs .(i) and (ii) , we get
` 2xx2^(2010) = 2a_(0) + a_(1) (omega + omega^(2)) + a_(2) (omega^(2) + omega^(4)) `
` + a_(3) (omega^(3) + omega^(6)) + a_(4) (omega^(4) + omega^(8))`
`+ a_(2) (omega^(5) + omega^(10)) + a_(5) (omega^(6) + omega^(12)) + ...`
` 2a_(0) - _(1) - a_(2) + 2a_(3)- 3a_(4) - a_(5) + 2a_(6) - ...`
`rArr 2^(2010) = a_(0) - 1/2a_(1)1/2a_(2) + a_(3) - 1/2 a_(4) - 1/2 a_(5) + _(6) -...`