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Sum of product of binomial coefficients

If a 0 , a ...

Question

If $a_{0}, a_{1}, a_{2}, . . . .$ be the coefficients in the expansion of ${(1 + x + x^{2})}^{n}$ in ascending powers of $x$ , also if $E_{1} = a_{0} + a_{3} + a_{6} + . . .,$ $E_{2} = a_{1} + a_{4} + a_{7} . . . .$ and $E_{3} = a_{2} + a_{5} + a_{8} + . . .;$ then:

A

E_{1} = E_{2} = E_{3}

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B

E_{n} = 3^{n - 1}

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C

Both A & B

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D

none of the above

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Solution

The correct option is C Both A & B

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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} + . . .

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

be the coefficients in the expansion of

(1 + x + x^{2})^{n}

in ascending powers of x, 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 + 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),

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

be the coefficients in the expansion of

{(1 + x + x^{2})}^{n}

in ascending powers of

x

a_{0} a_{1} - a_{1} a_{2} + a_{2} a_{3} - . . . . =

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