If 3-x2020-x20212022-a0-a1x-a2x2-a10xn- then the value of a0-12a1-12a2-a3-12a4-12a5-a6- iscorrect answer - 2- wrong answer - 0-50

Given : nbsp;(3+x^2020+x^2021)^2022 =a_0+a_1x+a_2x^2+⋯+a_10x^n Putting x=w, we get (3+w+w^2)^2022=a_0+a_1w+a_2w^2+⋯+a_10w^n ⇒ 2^2022=a_0+a_1w+a_2w^2+a_3+a_4 ...

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Byju's Answer

Standard XIII

Mathematics

Geometrical Representation of a Complex Number

If 3+x2020+x2...

Question

If $(3 + x^{2020} + x^{2021})^{2022} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{10} x^{n}$ , then the value of $a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} \dots$ is
(correct answer + 2, wrong answer - 0.50)

2^{2021}

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2^{2022}

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Solution

The correct option is C $2^{2022}$
Given : $(3 + x^{2020} + x^{2021})^{2022} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{10} x^{n}$

Putting $x = w$ , we get
$(3 + w + w^{2})^{2022} = a_{0} + a_{1} w + a_{2} w^{2} + \dots + a_{10} w^{n}$
$\Rightarrow 2^{2022} = a_{0} + a_{1} w + a_{2} w^{2} + a_{3} + a_{4} w + \dots$

Also, putting $x = w^{2}$ we get,
$2^{2022} = a_{0} + a_{1} w^{2} + a_{2} w + a_{3} + a_{4} w^{2} + \dots$

Adding both, we get
$\Rightarrow 2^{2023} = 2 a_{0} - a_{1} - a_{2} + 2 a_{3} - a_{4} - a_{5} + 2 a_{6} \dots ∴ a_{0} - \frac{a_{1}}{2} - \frac{a_{2}}{2} + a_{3} - \frac{a_{4}}{2} - \frac{a_{5}}{2} + a_{6} \dots = 2^{2022}$

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Similar questions

Q. If

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

is
(correct answer + 2, wrong answer - 0.50)

Q. 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

Q. 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} + . . . . . .

Q. If the expansion of

\frac{1}{(1 - a x) (1 - b x)} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} + \dots

, then

a_{n}

is (where

a \neq b, | a x |, | b x | < 1

)

y = x^{2} + \frac{1}{x^{2} + \frac{1}{\frac{1}{x^{2} + \dots}}}

, then

\frac{d y}{d x}

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If (3+x2020+x2021)2022=a0+a1x+a2x2+⋯+a10xn, then the value of a0−12a1−12a2+a3−12a4−12a5+a6⋯ is (correct answer + 2, wrong answer - 0.50)

If $(3 + x^{2020} + x^{2021})^{2022} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{10} x^{n}$ , then the value of $a_{0} - \frac{1}{2} a_{1} - \frac{1}{2} a_{2} + a_{3} - \frac{1}{2} a_{4} - \frac{1}{2} a_{5} + a_{6} \dots$ is
(correct answer + 2, wrong answer - 0.50)