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Quantitative Aptitude

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If 1+x-2x28...

Question

If $(1 + x - 2 x^{2})^{8} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{16} x^{16}$ then the sum $a_{1} + a_{3} + a_{5} + . . . . + a_{15}$ is equal to

A

- 2^{7}

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B

2^{7}

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C

2^{8}

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D

none of these

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Solution

The correct option is A $- 2^{7}$
Substituting x=1, we get
$0 = a_{0} + a_{1} + a_{2} + . . . a_{16}$
Substituting x=-1 we get
$2^{8} = a_{0} - a_{1} + a_{2} + . . . - a_{15} + a_{16}$
Subtracting ii from i, we get
$- 2^{8} = 2 [a_{1} + a_{3} + a_{5} + a_{7} + . . . a_{15}]$
Or
$a_{1} + a_{3} + a_{5} + a_{7} + . . . a_{15} = - 2^{7}$

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Similar questions

{(1 + x - 2 x^{2})}^{8} = a_{0} + a_{1} x + a_{2} x^{2} + \dots \dots + a_{16} x^{16},

sum : a_{1} + a_{3} + a_{5} + \dots + a_{15}

{(1 + x + x^{2})}^{8} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{16} x^{16}

x,

a_{5}

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

(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})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . a_{40} x^{40}

. Find the values of

a_{1} + a_{3} + a_{5} + . . . . a_{39}

{(1 + x - 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

and the value of

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

k =

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