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if ${(1 - x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$ then $a_{0} + a_{2} + a_{4} + \dots + a_{2 n}$ is equal to

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3n+12

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1−3n2

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Correct option is A

1−x+x2n=a0+a1x+a2x2+…+a2nx2n on putting X=1, we get(1−1+1)n=a0+a1+a2+…+a2n ⇒ 1=a0+a1+a2+…+a2n .....(i)Again, putting x= - 1, we get3n=a0−a1+a2−…+a2n3n+12=a0+a2+a4+…+a2n

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if 1−x+x2n=a0+a1x+a2x2+…+a2nx2n then a0+a2+a4+…+a2n is equal to

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if ${(1 - x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$ then $a_{0} + a_{2} + a_{4} + \dots + a_{2 n}$ is equal to