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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 theseA. `3^(2010)`B. `1`C. `2^(2010)`D. None of these |
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Answer» Correct Answer - C 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) "...."` |
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