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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...

If (1 – x + x2)n = a0 + a1 x + a2x2 + …..a2n x2n, then a0 + a2 + a4 +……a2n is equal to –

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

Putting x = 1,

1 = a0 + a1 + a2 + a3 +………a2n –––– (i)

Putting x = –1

3n = a0 – a1 + a2 – a3 +…….a2n –––– (ii)

Adding (i) & (ii);

3n+12=a0+a2+a4+.......a2n\frac{3^{n} + 1}{2} = a_{0} + a_{2} + a_{4} + .......a_{2n}