If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . .$ , then the value of $a_{2} + a_{4} + a_{6} + . . . + a_{12}$ will be

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Solution

The correct option is B 31
Substituting x=1, we get
$1 + a_{1} + a_{2} + a_{3} . . . = 0$ ...(i)
Substituting x=-1
$1 - a_{1} + a_{2} - a_{3} . . . = (- 2)^{6}$ ...(ii)
Adding i and ii, we get
$2 (1 + a_{2} + a_{4} + a_{6} . . .) = 64$
$1 + a_{2} + a_{4} + a_{6} . . . = 32$
$a_{2} + a_{4} + a_{6} . . . = 31$

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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_{3} x^{3} + \dots + a_{12} x^{12},

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

{(1 + x - 2 x^{2})}^{10} = 1 + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}

a_{2} + a_{4} + a_{6} + . . . + a_{20}

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