If ${(1 + x + x^{2})}^{8} = a_{0} + a_{1} x + . . . . . a_{16} x^{16}$ then $a_{0} - a_{2} + a_{4} - a_{6} + . . . . . + a_{16} =$

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Solution

The correct option is A 1
Putting $x = i$
$(1 + i + i^{2})^{8} = a_{0} + a_{1} i + a_{2} i^{2} + . . . . . + a_{1} 6 i^{1} 6$
$\Rightarrow (i)^{8} = a_{0} + a_{1} i - a_{2} + . . . . - - - - (1)$
putting $x = - i$
$\Rightarrow (1 - i + (- i)^{2})^{8} = a_{0} - a_{1} i + a_{2} (- i)^{2} + . . . . .$
$\Rightarrow (- i)^{8} = a_{0} - a_{1} i - a_{2} + . . . . . . . . . (2)$
equation $(1) +$ equation $(2) \Rightarrow i^{8} + i^{8} = 2 (a_{0} - a_{2} . . . . . . . .)$
$\Rightarrow 2 = 2 (a_{0} - a_{2} . . . . .)$
$\Rightarrow a_{0} - a_{2} . . . . . . .1 \Rightarrow (A)$

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