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If 1 + x - ...

Question

If ${(1 + x - 2 x^{2})}^{8} = a_{0} + a_{1} x + a_{2} x^{2} + \dots \dots + a_{16} x^{16},$ then the sum $sum : a_{1} + a_{3} + a_{5} + \dots + a_{15}$ is :

A

2^{8}

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B

- 2^{8}

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C

2^{7}

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D

- 2^{7}

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Solution

The correct option is C $- 2^{7}$
$(1 + x - 2 x^{2})^{8} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{16} x^{16} p e t x = 1, w e g e t (1 + 1 - 2)^{8} = a_{0} + a_{1} + a_{2} + . . . . . . . . . + a_{16} = 0 t h e n p u t x = - 1, w e g e t (1 - 1 - 2 (- 1)^{2})^{8} = a_{0} - a_{1} + a_{2} - a_{3} + . . . . . . . + a_{16} = 2^{8} s u b t r a c t b o t h n e w e q u a t i o n, w e g e t 0 - 2^{8} = 2 (a_{1} + a_{3} + . . . . . . + a_{15}) ∴ a_{1} + a_{3} + . . . . . . + a_{15} = - 2^{7}$

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Similar questions

(1 + x - 2 x^{2})^{8} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{16} x^{16}

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

(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}

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