Let 1-x1-x-x- 21-x-x- 2-x- 3 - - -1-x-x- 2- -x -30 -a- 0-a- 1 x-a- 2 - 12- -a-465- x -465then sum of a0-a2-a4- - -a464 is

The correct option is B (31)!/2Putting x = 1 a_0 + a_1 + a_2 + .......+a_464 = (31)!.....(i) Putting x = 1 a_0 a_1 + a_2 ....... +a_464= 0.....(ii) Adding ...

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Byju's Answer

Standard XI

Mathematics

Definition of Functions

Let 1+x1+x+x∧...

Question

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

then sum of $a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{464}$ is

(31)!

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\frac{(31)!}{2}

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(30)!

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\frac{(60)!}{2}

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Solution

The correct option is B $\frac{(31)!}{2}$
Putting $x = 1$
$a_{0} + a_{1} + a_{2} + . . . . . . . + a_{464} = (31)! . . . . . (i)$
Putting $x = - 1$
$a_{0} - a_{1} + a_{2} - . . . . . . . + a_{464} = 0..... (i i)$
Adding both the equation,
$a_{0} + a_{2} + a_{4} + . . . . . . = \frac{(31)!}{2}$

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Similar questions

Q. Let

(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}

then sum of

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

Q. If

\frac{1}{x (x + 1) (x + 2) . . . (x + n)} = \frac{A_{0}}{x} + \frac{A_{1}}{x + 1} + \frac{A_{2}}{x + 2} + . . . . . + \frac{A_{n}}{x + n}

A_{r} =

Q. If

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

, then the value of

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

Q. If

\frac{1}{x (x + 1) (x + 2) \dots (x + n)} = \frac{A_{0}}{x} + \frac{A_{1}}{x + 1} + \frac{A_{2}}{x + 2} + \dots + \frac{A_{n}}{x + n}

, then

A_{r}

is equal to

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & 3 x - 1 & - x + 3 2 x + 1 & 2 + x^{2} & x^{3} - 3 x - 3 & x^{2} + 4 & 3 x \end{matrix} ∣ ∣ ∣ ∣ ∣ = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{7} x^{7}

, then the value of

a_{0}

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Let (1+x)(1+x+x2)(1+x+x2+x3)......(1+x+x2+....+x30)=a0+a1x+a2x2+........+a465x465 then sum of a0+a2+a4+........+a464 is

The correct option is B (31)!2Putting x=1 a0+a1+a2+.......+a464=(31)!.....(i) Putting x=−1 a0−a1+a2−.......+a464=0.....(ii) Adding both the equation, a0+a2+a4+......=(31)!2

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

then sum of $a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{464}$ is

The correct option is B $\frac{(31)!}{2}$
Putting $x = 1$
$a_{0} + a_{1} + a_{2} + . . . . . . . + a_{464} = (31)! . . . . . (i)$
Putting $x = - 1$
$a_{0} - a_{1} + a_{2} - . . . . . . . + a_{464} = 0..... (i i)$
Adding both the equation,
$a_{0} + a_{2} + a_{4} + . . . . . . = \frac{(31)!}{2}$