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Equal Vectors

If 1+x+x2n=...

Question

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

A

a_{1} + a_{4} + a_{7} . . . . . .

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B

a_{2} + a_{5} + a_{8} . . . . . .

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C

3^{n - 1}

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D

2^{n}

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Solution

The correct options are
A $a_{2} + a_{5} + a_{8} . . . . . .$
B $a_{1} + a_{4} + a_{7} . . . . . .$
C $3^{n - 1}$
${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}$
Put $x = 0$ , $1 = a_{0}$
Put $x = 1$ , $3^{n} = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + . . . . . + a_{2 n}$
$3^{n} = (a_{0} + a_{3} + a_{6} + a_{9} + . . . .) + (a_{1} + a_{4} + a_{7} + . . . .) + (a_{2} + a_{5} + . . .)$ ... $(i)$
Put $x = ω$ ,
${(1 + ω + ω^{2})}^{n} = a_{0} + a_{1} ω + a_{2} ω^{2} + . . . . + a_{2 n} ω^{2 n}$
$\Rightarrow 0 = (a_{0} + a_{3} + a_{6} + a_{9} + . . . .) + (a_{1} + a_{4} + a_{7} + . . . .) ω + (a_{2} + a_{5} + . . .) ω^{2}$ ... $(i i)$

Put $x = ω^{2}$ ,
${(1 + ω^{2} + ω^{4})}^{n} = a_{0} + a_{1} ω^{2} + a_{2} ω^{4} + . . . . + a_{2 n} (ω^{2})^{2 n}$
$\Rightarrow 0 = (a_{0} + a_{3} + a_{6} + a_{9} + . . . .) + (a_{1} + a_{4} + a_{7} + . . . .) ω^{2} + (a_{2} + a_{5} + . . .) ω^{4}$ ... $(i i i)$

Adding $(i), (i i), ((i i i)$
$3^{n} = 3 (a_{0} + a_{3} + a_{6} + . . .)$
$\Rightarrow a_{0} + a_{3} + a_{6} + . . . = 3^{n - 1}$ option $C$
Putting this in $(i), (i i)$ , options $A$ and $B$ also correct.

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Similar questions

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

then prove that

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

Q. If the expansion of

(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}

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)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{10}

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

{(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}

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

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