If $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a^{2 n} x^{2 n}, t h e n a_{0} + a_{3} + a_{6} + . . . . . =$

3ⁿ

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3^n-1

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3^n-2

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Solution

The correct option is B
3^n-1

Put $x = 1, ω, ω^{2}$ then
$x = 1 \Rightarrow (3)^{n} = a_{0} + a_{1} + a_{2} + a_{3} + . . . .$ ,
$x = ω \Rightarrow (1 + ω + ω^{2})^{n} = a_{0} + a_{1} ω + a_{2} ω^{2} + a_{3} + . . . .$ ,
$x = ω^{2} \Rightarrow (1 + ω + ω^{2})^{n} = a_{0} + a_{1} ω^{2} + a_{2} ω + a_{3} + . . . .$ ,
Adding all we get $3^{n} = 3 (a_{0} + a_{1} + a_{2} + a_{3} + . . . .)$

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Similar questions

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

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

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

a_{0} + a_{3} + a_{6} . . . . . .

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

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

(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^{2} + . . . + a_{2 n} x^{2 n}

a_{0} + a_{2} + a_{4} + . . . + a_{2 n}

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