NIMCET 2017 — Mathematics PYQ

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Question

NIMCET 2017 — Mathematics PYQ

NIMCET | Mathematics | 2017

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?

Choose the correct answer:

Explanation

Calculation:

Given,
$(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$

Put, $x = 1$ ,

$(1 - 1 + 1^{2})^{n} = a_{0} + a_{1} + a_{2} + \dots + a_{2 n}$

$\Rightarrow 1 = a_{0} + a_{1} + a_{2} + \dots + a_{2 n}$ ......(1)

Put, $x = - 1$ ,

$(1 + 1 + (- 1)^{2})^{n} = a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{2 n}$

$\Rightarrow 3^{n} = a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{2 n}$ ......(2)

Adding (1) and (2),

$2 (a_{0} + a_{2} + a_{4} + \dots + a_{2 n}) = 3^{n} + 1$

$\Rightarrow a_{0} + a_{2} + a_{4} + \dots + a_{2 n} = \frac{3 ^{n} + 1}{2}$

Hence, option (1) is correct.