1959 AMC 12 第 39 题

Question

设

S

为下列数列前九项之和：

\begin{gathered} x+a,\quad x^2+2a,\\ x^3+3a,\quad\ldots \end{gathered}\text{。}

则

S

等于：

Let

S

be the sum of the first nine terms of the sequence

\begin{gathered} x+a,\quad x^2+2a,\\ x^3+3a,\quad\ldots. \end{gathered}

Then

S

equals:

\dfrac{50a+x+x^8}{x+1}

50a-\dfrac{x+x^{10}}{x-1}

\dfrac{x^9-1}{x+1}+45a

\dfrac{x^{10}-x}{x-1}+45a

\dfrac{x^{11}-x}{x-1}+45a

将

x

的幂与

a

的倍数分开

Separate the powers of

x

from the multiples of

a

对

x+x^2+\cdots+x^9

使用等比数列求和公式

Use the geometric-series sum for

x+x^2+\cdots+x^9

将前九项相加，得

\begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a \end{aligned}\text{。}

因此

S=\frac{x^{10}-x}{x-1}+45a\text{。}

当

x=1

时按连续性理解此表达式，此时两种形式都等于

9+45a

。

因此，正确答案是 D。

Adding the first nine terms gives

\begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a. \end{aligned}

Therefore

S=\frac{x^{10}-x}{x-1}+45a.

The expression is understood by continuity at

x=1,

where both forms equal

9+45a.

Thus, the correct answer is D.

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