If $(1 + x + x^{2})^{n} = a_{0} + a_{1} + a_{2} x^{2 n}$ , then $a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{2 n} x^{2 n},$ is equal to :

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

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\frac{3^{n} - 1}{2}

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\frac{1 - 3^{n}}{2}

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

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Solution

The correct option is B $\frac{3^{n} + 1}{2}$
${(1 + x + x^{2})}^{4} = a_{0} + a_{1} + a_{2} x^{2 n} + . . . . . . . . . x^{4} + . . . . . . . . a_{2 x} x^{2 n} \to (i)$

$P u t x = 1 i n e q u a t i o n (i) 3^{x} = a_{0} + a_{1} + a_{2} + . . . . . . . . . . . . . . . . . . . . \to (i i)$

$P u t x = - 1 i n e q u a t i o n (i) 1 = a_{0} - a_{1} + a_{2} - a_{3} + . . . . . . . . . . . . . . \to (i i i)$

$O n a d d i n g e q u a t i o n (i i) a n d (i i i) 3^{n} + 1 = 2 (a_{0} + a_{2} + a_{4} + . . . . . . . . .) \frac{3^{n} + 1}{2} = a_{0} + a_{2} + a_{4} + . . . . . . . . . . . .$

Hence, this is the answer.

Suggest Corrections

Similar questions

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

\frac{3^{n} - 1}{2}

\frac{1 - 3^{n}}{2}

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

(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}

E_{1} = a_{0} + a_{3} + a_{6} + . . .

E_{2} = a_{1} + a_{4} + a_{7} + . . .

E_{3} = a_{2} + a_{5} + a_{8} + . . .

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

{(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2} a_{r} x^{r} t h e n a_{0} + a_{4} \dots + a_{2 n}

(1 + x + x^{2})^{n}

a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{r} x^{r} + . . . . . + a_{2 n} x^{2 n}

a_{0} + a_{3} + a_{6} + . . . . . = a_{1} + a_{4} + a_{7} + . . . . . = a_{2} + a_{5} + a_{8} + . . . . = 3^{n - 1}

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