AMC12 2008 A

AMC12 2008 A · Q19

AMC12 2008 A · Q19. It mainly tests Polynomials, Basic counting (rules of product/sum).

In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$?

在展开式 \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] 中，$x^{28}$ 的系数是多少？

(A) 195 195

(B) 196 196

(D) 378 378

(E) 405 405

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$ and $B = \left(1 + x + x^2 + \cdots + x^{27}\right)$. We are expanding $A \cdot A \cdot B$. Since there are $15$ terms in $A$, there are $15^2 = 225$ ways to choose one term from each $A$. The product of the selected terms is $x^n$ for some integer $n$ between $0$ and $28$ inclusive. For each $n \neq 0$, there is one and only one $x^{28 - n}$ in $B$. For example, if I choose $x^2$ from $A$ , then there is exactly one power of $x$ in $B$ that I can choose; in this case, it would be $x^{24}$. Since there is only one way to choose one term from each $A$ to get a product of $x^0$, there are $225 - 1 = 224$ ways to choose one term from each $A$ and one term from $B$ to get a product of $x^{28}$. Thus the coefficient of the $x^{28}$ term is $224 \Rightarrow \boxed{C}$.

设 $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$，$B = \left(1 + x + x^2 + \cdots + x^{27}\right)$。我们在展开 $A \cdot A \cdot B$。由于 $A$ 有 $15$ 项，从两个 $A$ 中各选一项共有 $15^2 = 225$ 种方式。所选两项的乘积为 $x^n$，其中整数 $n$ 介于 $0$ 到 $28$（含）之间。对每个 $n \neq 0$，在 $B$ 中恰有一个 $x^{28 - n}$。例如，若从 $A$ 中选了 $x^2$，则在 $B$ 中恰有一个可选的幂次使总次数为 $28$，此时为 $x^{24}$。由于从两个 $A$ 中选出乘积为 $x^0$ 的方式只有一种，因此从两个 $A$ 与一个 $B$ 中选项使乘积为 $x^{28}$ 的方式共有 $225 - 1 = 224$ 种。故 $x^{28}$ 项的系数为 $224 \Rightarrow \boxed{C}$。

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