Question

数列 $1440$ ， $1716$ ， $1848, \ldots$ 的每一项由两个等差数列的对应项相乘得到。求这个数列的第八项。

Find the eighth term of the sequence $1440,$ $1716,$ $1848, \ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

答案

答案：348

知识点：等差数列二次方程方程组

难度评级：2340

解答：

等差数列的第 $n$ 项是 $n$ 的线性函数，因此两个等差数列对应项的乘积是一个二次式 $t_n = an^2 + bn + c$ 。将给出的三项依次编号为 $n = 0, 1, 2$ ，则 $\begin{aligned} c &= 1440, \\ a + b + c &= 1716, \\ 4a + 2b + c &= 1848, \end{aligned}$ 得到 $a + b = 276$ 和 $2a + b = 204$ ，所以 $a = -72$ ， $b = 348$ ， $c = 1440$ 。

第八项为 $t_7 = -72 \cdot 49$ $+ 348 \cdot 7$ $+ 1440 = 348$ 。（确实有 $t_n = (180 - 24n)(8 + 3n)$ ，它是两个等差数列的乘积，并符合给出的项。）

The $n$ th term of an arithmetic sequence is linear in $n,$ so the product of corresponding terms of two arithmetic sequences is a quadratic $t_n = an^2 + bn + c.$ Indexing the given terms by $n = 0, 1, 2:$ $\begin{aligned} c &= 1440, \\ a + b + c &= 1716, \\ 4a + 2b + c &= 1848, \end{aligned}$ which give $a + b = 276$ and $2a + b = 204,$ so $a = -72,$ $b = 348,$ $c = 1440.$

The eighth term is $t_7 = -72 \cdot 49$ $+ 348 \cdot 7$ $+ 1440 = 348.$ (Indeed $t_n = (180 - 24n)(8 + 3n),$ a product of two arithmetic sequences matching the given terms.)

2003 AIME II 第 8 题

答案：348

难度评级：2340

解答：

其他年份的第 8 题