Let (a + bx + cx^2)^{10} = sum_{i=0}^{20} p_i | vedclass

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(B) Given $(a + bx + cx^2)^{10} = \sum_{i=0}^{20} p_i x^i$.The coefficient of $x^1$ is $p_1 = 10 	imes a^9 	imes b = 20$.Thus,$a^9 b = 2$. Since $a,

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$12$

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(B) Given $(a + bx + cx^2)^{10} = \sum_{i=0}^{20} p_i x^i$.The coefficient of $x^1$ is $p_1 = 10 	imes a^9 	imes b = 20$.Thus,$a^9 b = 2$. Since $a, b \in N$,we must have $a = 1$ and $b = 2$.The coefficient of $x^2$ is $p_2 = \binom{10}{1} a^9 c + \binom{10}{2} a^8 b^2 = 210$.Substituting $a = 1$ and $b = 2$:$10(1)^9 c + 45(1)^8 (2)^2 = 210$.$10c + 45(4) = 210$.$10c + 180 = 210$.$10c = 30$,so $c = 3$.Finally,$2(a + b + c) = 2(1 + 2 + 3) = 2(6) = 12$.

Let $(a + bx + cx^2)^{10} = \sum_{i=0}^{20} p_i x^i$,where $a, b, c \in N$. If $p_1 = 20$ and $p_2 = 210$,then $2(a + b + c)$ is equal to

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