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

21

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11

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31

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None of the above

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Solution

The correct option is C $31$
${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{12} x^{12}$

Putting $x = 1$ , we get $0 = 1 + a_{1} + a_{2} + . . . + a_{12}$ ....(1)

Putting $x = - 1$ , we get $64 = 1 - a_{1} + a_{2} - . . . + a_{12}$ ...(2)

Adding (1) and (2), we get

$64 = 2 (1 + a_{2} + a_{4} + . . .)$

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

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Similar questions

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

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

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