Question

设 $S_n$ 是一个公差为 $2$ 的等差数列的前 $n$ 项和。商 $\dfrac{S_{3n}}{S_n}$ 与 $n$ 无关。求 $S_{20}$ 。

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2.$ The quotient $\dfrac{S_{3n}}{S_n}$ does not depend on $n.$ What is $S_{20}?$

$340$

$360$

$380$

$400$

$420$

答案：D

知识点：等差数列求和

难度评级：1820

解答：

设数列为 $n$ 。令 $a$ 为 $a+2n$ 前一项的值，则。 $\begin{aligned}S_n&=\sum_{i=1}^n(a+2i)\\&=n(a+n+1).\end{aligned}$

因此 $\begin{aligned} \frac{S_{3n}}{S_n}&=\frac{3(a+3n+1)}{a+n+1}\\ &=9-\frac{6(a+1)}{a+n+1}. \end{aligned}$

于是需要选择 $6(a+1)$ ，使这个值为常数。若这个值为常数，则它减去 $a=-1$ 后也为常数：分母随 $n$ 变化，因此分子必须为 $0$ 。

于是，。 $S_{20}=20(-1+20+1)=400.$

所以答案是 D。

Write the $n$ th term as $a+2n,$ so the term before the first term is $a.$ Then $\begin{aligned}S_n&=\sum_{i=1}^n(a+2i)\\&=n(a+n+1).\end{aligned}$

Hence $\begin{aligned} \frac{S_{3n}}{S_n}&=\frac{3(a+3n+1)}{a+n+1}\\ &=9-\frac{6(a+1)}{a+n+1}. \end{aligned}$

For this expression to be independent of $n,$ its numerator $6(a+1)$ in the final fraction must be $0.$ Thus $a=-1.$

Therefore, $S_{20}=20(-1+20+1)=400.$

Thus, the answer is D .

2022 AMC 10B 第 15 题

答案：D

难度评级：1820

解答：

其他年份的第 15 题