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Binomial Coefficients

Let 1 + x +...

Question

Let $((1 + x) + x^{2})^{9} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{18} x^{18}$ . Then

A

a_{0} + a_{2} + . . . . . + a_{18} = a_{1} + a_{3} + . . . . . + a_{17}

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B

a_{0} + a_{2} + . . . . . + a_{18}

is even

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C

a_{0} + a_{2} + . . . . . + a_{18}

is divisible by

9

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D

a_{0} + a_{2} + . . . . . + a_{18}

is divisible by

3

but not by

9

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Solution

The correct option is A $a_{0} + a_{2} + . . . . . + a_{18}$ is even
Given : $((1 + x) + x^{2})^{9} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{18} x^{18}$ ..... $(i)$
Put $x = 1$ in $(i)$ , we get
$(1 + 1 + 1)^{9} = a_{0} + a_{1} + a_{2} + . . . . . + a_{18}$
$3^{9} = a_{0} + a_{1} + a_{2} + . . . . . + a_{18}$ .... $(i i)$
Put $x = - 1$ in $(i)$ , we get
$(1 - 1 + 1)^{9} = a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - . . . . . - a_{17} + a_{18}$
$1 = a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - . . . . . - a_{17} + a_{18}$ ..... $(i i i)$
Adding $(i i)$ and $(i i i)$ , we get
$3^{9} + 1 = a_{0} + a_{1} + . . . . + a_{18} + a_{0} - a_{1} + a_{2} - . . . . . - a_{17} + a_{18}$
$= 2 a_{0} + 2 a_{2} + 2 a_{4} + . . . . + 2 a_{18}$
$⟹ 2 (a_{0} + a_{2} + a_{4} + . . . . + a_{18}) = 3^{9} + 1$
$⟹ a_{0} + a_{2} + . . . . . + a_{18} = \frac{3^{9} + 1}{2} \to e v e n$
Hence, $a_{0} + a_{2} + a_{4} + . . . . + a_{18}$ is even.

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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},

find the values of
(i)

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

Q. Assertion :Let

(1 + x^{2}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{36} x^{36}

a_{0} + a_{3} + a_{6} + . . . + a_{36} = \frac{2}{3} (2^{35} + 1)

a_{0} + a_{1} + a_{2} + . . . + a_{36} = 2^{36}

a_{0} + a_{2} + a_{4} + . . . + a_{36} = 2^{35}

(1 + x + x^{2})^{2014} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . + a_{4028} x^{4028}

A = a_{0} - a_{3} + a_{6} - . . . . . + a_{4026}

,

B = a_{1} - a_{2} + a_{7} - . . . . . + a_{4027}

,

C = a_{2} - a_{5} + a_{8} - . . . . . + a_{4028}

$(1 + x + 2 x^{2})^{20}$ = $a_{0} + a_{1} x + a_{2} x^{2}$ ................... $a_{40} x^{40}$ , If $a_{0} + a_{1} + a_{2} + a_{3} . . . . . . . . . . . . . . . . . . a_{40} = 2^{a}$

$a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . . . . . . . . . . . . . . a_{40} = 2^{b}$ . Find the value of a + b

\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

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