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

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

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

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

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

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Solution

The correct option is A $\frac{3^{n} + 1}{2}$
We have,
$a_{1} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . . . . + a_{2 n} x^{2 n} = (1 - x + x^{2})^{n}$
On putting $x = 1$ and $- 1$ respectively we get
$(a_{0} + a_{2} + a_{4} + . . . .) + (a_{1} + a_{3} + a_{5} + . . . .) = 1$ ..........(i)
and $(a_{0} + a_{2} + a_{4} + . . . . .) - (a_{1} + a_{3} + a_{5} + . . . . .) = 3^{n}$ ............(ii)
On adding Eqs. (i) and (ii), we get
$2 (a_{0} + a_{2} + a_{4} + . . . .) = 3^{n} + 1$
$a_{0} + a_{2} + a_{4} + . . . . = \frac{3^{n} + 1}{2}$

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{(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_{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^{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}

(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),

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