If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ , then $a_{0} + a_{3} + a_{6} + . . . =$

3^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3^{n - 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3^{n - 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3^{n - 1}$
Putting $x = 1, ω, ω^{2},$ respectively,

$3^{n} = a_{0} + a_{1} + a_{2} + . . . + a_{2 n}$

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

${(1 + ω^{2} + ω^{4})}^{n} = a_{0} + a_{1} ω^{2} + a_{2} ω^{4} + . . + a_{2 n} ω^{4 n}$

Adding these,

$3^{n} + {(1 + ω + ω^{2})}^{n} + {(1 + ω^{2} + ω^{4})}^{n}$

$= 3 a_{0} + a_{1} (1 + ω + ω^{2}) + a_{2} (1 + ω^{2} + ω^{4}) + a_{3} (1 + ω^{3} + ω^{6}) + . . .$

$\Rightarrow 3^{n} + 0^{n} + 0^{n} = 3 a_{0} + 3 a_{3} + 3 a_{6} + . . .$

$\Rightarrow 3^{n - 1} = a_{0} + a_{3} + a_{6} + . . .$

Suggest Corrections

Similar questions

{(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^{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),

\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^{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}

Join BYJU'S Learning Program