If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{12} x^{12}$ , then the expression $a_{2} + a_{4} + a_{6} + . . . . . . + a_{12}$ has the value

32

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63

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64

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31

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Solution

The correct option is D $31$
Given $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{12} x^{12}$
Put $x = 1$ in the given equation
$0 = 1 + a_{1} + a_{2} + a_{3} + . . . . + a_{12}$ ....(1)
Put $x = - 1$ in the given equation
$(- 2)^{6} = 1 - a_{1} + a_{2} - a_{3} + . . . . + a_{12}$ .....(2)
Adding (1) and (2), we get
$0 + (- 2)^{6} = (1 + a_{1} + a_{2} + . . . . + a_{12}) + (1 - a_{1} + a_{2} - a_{3} . . . . . . + a_{12})$
$64 = 2 (1 + a_{2} + a_{4} + . . . . . + a_{12})$
$∴ 1 + a_{2} + a_{4} + . . . . . . + a_{12} = 32$
$\Rightarrow a_{2} + a_{4} + . . . . . . + a_{12} = 31$

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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}

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}

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

x^{4}

2 x^{2})^{6}

1 + a_{1} x + a_{2} x^{2} + . . . . + a_{1} 2 x^{12}

a_{2} + a_{4} + a_{6} + . . . + a_{12} = 31

{(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{12} x^{12},

a_{2} + a_{4} + a_{6} + \dots + a_{12}

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