If 1-x-x21-x-1-x2-2-x3-3- a 0 - a 1 x- a 2 x 2 - a 3 x 3 - a 4 x 4 - then-

The correct option is A a _ 2 = 12 ( 1+x+x^ 2 ) ( 1 x 1! + x^ 2 2! x^ 3 3! + ) = a _ 0 + a _ 1 x+ a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ ...

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Combination

If 1+x+x21-...

Question

If $(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .$ then,

a_{1} = - 1

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a_{2} = \frac{1}{2}

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a_{3} = \frac{1}{3}

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a_{4} = - \frac{1}{2}

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(1 + x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) . . . . . . (1 + x + x^{2} + . . . . + x^{30}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{465} x^{465}

a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{464}

(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}

a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0} = 0

a_{1}, i = 1, 2, 3, 4

{(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}

x + 1

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

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