If 3-x2008-x20092010-a0-a1x-a2x2-anxn-then the value ofa0-12a1-12a2-a3-12a4-12a5-a6- -upto n terms is

Put x=w,,we get (3+w+w^2)^2010=a_0+a_1w+a_2w^2+a_3ω^3+⋯ ⇒ 2^2010=a_0+a_1w+a_2w^2+a_3+ ⋯ …(1) Put x=w^2,we get (3+w^2+w^4)^2010=a_0+a_1w^2+a_2w+a_3ω^6+⋯ ⇒ 2^20 ...

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Byju's Answer

Standard XII

Mathematics

Square Root of a Complex Number

If 3+x2008+x2...

Question

If $(3 + x^{2008} + x^{2009})^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ ,
then the value of
$a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} - \dots$ upto $n$ terms is

3^{2010}

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1

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2^{2010}

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4^{2010}

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Solution

The correct option is C $2^{2010}$
Put $x = w,$ ,we get
$(3 + w + w^{2})^{2010} = a_{0} + a_{1} w + a_{2} w^{2} + a_{3} ω^{3} + \dots$
$\Rightarrow 2^{2010} = a_{0} + a_{1} w + a_{2} w^{2} + a_{3} + \dots \dots (1)$
Put $x = w^{2}$ ,we get
$(3 + w^{2} + w^{4})^{2010} = a_{0} + a_{1} w^{2} + a_{2} w + a_{3} ω^{6} + \dots$
$\Rightarrow 2^{2010} = a_{0} + a_{1} w^{2} + a_{2} w + + a_{3} + \dots \dots (2)$
Adding $(1), (2)$ , we get
$\Rightarrow 2 \times 2^{2010} = 2 a_{0} - a_{1} - a_{2} + 2 a_{3} - a_{4} - a_{5} + 2 a_{6} + \dots$
$\Rightarrow 2^{2010} = a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} \dots$

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If (3+x2008+x2009)2010=a0+a1x+a2x2+…+anxn, then the value of a0−12a1−12a2+a3−12a4−12a5+a6− ⋯upto n terms is

If $(3 + x^{2008} + x^{2009})^{2010} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ ,
then the value of
$a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} - \dots$ upto $n$ terms is