If x 2006 - x 2008 -2 2010 - a 0 - a 1 x- a 2 x 2 - a 0 x n -1then value of a 0 - 1 2 a 1 - 1 2 a 2 - a 3 - 1 2 a 4 - 1 2 a 5 - a 6 - is

The correct options are A less than 2 B greater than 0 Substituting x=w we get nbsp; (w+w^2+2)^2010 =(1) =a_0+a_1w+a_2w^2+a_3+a_ ...

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Algebra of Derivatives

If x 2006 +...

Question

If
${(x^{2006} + x^{2008} + 2)}^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a_{0} x^{n} . . . . . . . . . (1)$
then value of
$a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} + . . . . . .$ is

less than

2

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greater than

0

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equals

2

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none of these

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Similar questions

(3 + x^{2008} + x^{2009})^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} - \dots

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(3 + x^{2020} + x^{2021})^{2022} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{10} x^{n}

a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} \dots

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . . . . . + a_{2 n} x^{2 n}

a_{0} + a_{3} + a_{6} + . . . . .

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a_{0} + a_{3} + a_{6} + . . . =

{(x^{2006} + \frac{1}{x^{2006}} + 2)}^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . a_{n} x^{n}

2 a_{0} - a_{1} - a_{2} + 2 a_{3} - a_{4} - a_{5} + 2 a_{6} - . . .

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