Let (1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) d | vedclass

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(B) Let $P(x) = (1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{30}) = \sum_{k=0}^{465} a_k x^k$.Note that each term $(1 + x +

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$\frac{(31)!}{2}$

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(B) Let $P(x) = (1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{30}) = \sum_{k=0}^{465} a_k x^k$.Note that each term $(1 + x + x^2 + \dots + x^n) = \frac{1-x^{n+1}}{1-x}$.Substituting $x=1$ into the product,each factor $(1 + 1 + 1^2 + \dots + 1^n) = (n+1)$.Thus,$P(1) = 1 	imes 2 	imes 3 	imes \dots 	imes 31 = 31!$.So,$a_0 + a_1 + a_2 + \dots + a_{465} = 31!$ $(1)$.Substituting $x=-1$ into the product,the first factor $(1 + (-1)) = 0$.Thus,$P(-1) = 0$.So,$a_0 - a_1 + a_2 - a_3 + \dots = 0$ $(2)$.Adding $(1)$ and $(2)$,we get $2(a_0 + a_2 + a_4 + \dots) = 31!$.Therefore,$a_0 + a_2 + a_4 + \dots = \frac{31!}{2}$.

Let $(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{30}) = a_0 + a_1x + a_2x^2 + \dots + a_{465}x^{465}$. Then the sum $a_0 + a_2 + a_4 + \dots$ is equal to:

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