If ${(1 + x + x^{2})}^{n} = 2 n \sum r = 0 a_{r} x^{r} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a^{2 n} x^{2 n}$ and
$P = a_{0} + a_{3} + a_{6} + . . .$
$Q = a_{1} + a_{4} + a_{7} + . . .$
$R = a_{2} + a_{5} + a_{8} + . . .$
then the set of values of $P, Q, R$ are respectively equals

(1, 1, 1)

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(3^{n}, 3^{n}, 3^{n})

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(3^{n + 1}, 3^{n + 1}, 3^{n + 1})

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(3^{n - 1}, 3^{n - 1}, 3^{n - 1})

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Solution

The correct option is C $(3^{n - 1}, 3^{n - 1}, 3^{n - 1})$
Sum of all coefficients will be $3^{n}$
$= 3 (3^{n - 1})$
Now
The sum of coefficients of $a_{0} + a_{3} + a_{6} . . . .$
will be equal to $a_{1} + a_{4} + a_{7} + . . . .$ which will be further equal to
$a_{2} + a_{5} + a_{8} + . . .$
$=$ sum of all coefficients/3
$= \frac{3 (3^{n - 1})}{3} = 3^{n - 1}$

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Similar questions

(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),

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

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

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

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

a_{0}, a_{1}, a_{2}, . . . . .

(1 + x + 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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