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Question: If (1–x+x<sup>2</sup>)<sup>n</sup>=a<sub>0</sub>+a<sub>1</sub>x+a<sub>2</sub>x<sup>2</sup> + …. + a<...

If (1–x+x2)n=a0+a1x+a2x2 + …. + a2nx2n then a0 + a2 + a4 + …. +a2n equals

A

3n+12\frac{3^{n} + 1}{2}

B

3n−12\frac{3^{n} - 1}{2}

C

1−3n2\frac{1 - 3^{n}}{2}

D

3n+123^{n} + \frac{1}{2}

Answer

3n+12\frac{3^{n} + 1}{2}

Explanation

Solution

Put x = 1 ⇒ 1 = a0 + a1 + a2 + + a2n.

Put x = −1 ⇒ 3n = a0 − a1 + a2 − a3 + + a2n.

Adding, 3n + 1 = 2 (a0 + a2 + a4 + + a2n)