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Roots of Quadratic Equation

if 1+x-2x2...

Question

if ${(1 + x - 2 x^{2})}^{10} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}$ , then the value of $a_{2} + a_{4} + a_{6} + . . . + a_{20}$

A

1,024

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B

512

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C

1,023

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D

511

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Solution

The correct option is D 511
$(1 + x - 2 x^{2})^{10} = 1 + a_{1} (x) + a_{2} x^{2} . . . a_{20} x^{20}$
Substituting, $x = 1$ we get
$0 = 1 + a_{1} + a_{2} + a_{3} . . . a_{20}$ ...(i)
Substituting $x = - 1$ , we get
$(2)^{10} = 1 - a_{1} + a_{2} - a_{3} . . . + a_{20}$ ...(ii)
Adding i and ii we get
$2^{10} = 2 (1 + a_{2} + a_{4} + a_{6} + . . . a_{20})$
$2^{9} = 1 + a_{2} + a_{4} + a_{6} + . . . a_{20}$
$2^{9} - 1 = a_{2} + a_{4} + a_{6} + . . . a_{20}$
Hence
$a_{2} + a_{4} + a_{6} + . . . a_{20}$
$= 2^{9} - 1$
$= 512 - 1$
$= 511$

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Similar questions

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

a_{2} + a_{4} + a_{6} + . . . . + a_{12} =

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

, then the expression

a_{2} + a_{4} + a_{6} + . . . + a_{12}

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{12} x^{12}

, then the expression

a_{2} + a_{4} + a_{6} + . . . . . . + a_{12}

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

, then value of

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

(1 - x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

\frac{a_{2}}{2^{2}} + \frac{a_{4}}{2^{4}} + \frac{a_{6}}{2^{6}} + . . . . + \frac{a_{12}}{2^{12}}

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