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Let 1+ x1 + 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

A

(31)!

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B

\frac{(31)!}{2}

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C

(30)!

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D

\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

(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} + . . . . .

(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} + . . .

(1 + x + x^{2} + x^{3} + \dots \dots . x^{p})^{n} = a_{0} + a_{1^{X}} + a_{2} x^{2} + \dots . . + a_{n p} x^{n p}

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

(1 \times x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) . . . . . (1 + x + x^{2} + x^{3} + . . . . . . + x^{n}) \equiv

a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . . . a_{m} x^{m}

m \sum r = 0 a_{r}

has the value equal to

(n + k)!

k

(1 + x + x^{2} + x^{3} + x^{4})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{40} x^{40}

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