Question

一个数阵从顶行的 $-1, 3, 1$ 开始构造。每一对相邻数字相加，得到下一行中的一个数字。每一行的开头和结尾分别是 $-1$ 和 $1$ ，前三行如下图所示。

如果这个过程继续下去，某一行的和将为 $12{,}288$ 。在那一行中，从左数第三个数是多少？

An array of numbers is constructed beginning with the numbers $-1, 3, 1$ in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row will begin and end with the numbers $-1$ and $1,$ respectively. The first three rows are shown below.

If the process continues, one of the rows will sum to $12{,}288.$ In that row, what is the third number from the left?

$-29$

$-21$

$-14$

$-8$

$-3$

答案：A

知识点：递推三角形数找规律

难度评级：1910

解答：

每个内部项会贡献给下一行的两个项，而端点 $-1$ 和 $1$ 在总和中相互抵消。因此每一行的总和都是上一行的两倍。顶行总和为 $3$ 且 $12{,}288 = 3 \cdot 2^{12}$ 所以这是第 $12$ 行（顶行算第 $0$ 行）。从左侧追踪对角线。第二条对角线为 $3, 2, 1, 0, -1, \ldots$ 每行减少 $1$ 。第三条对角线是这些数的累加：当 $n \ge 4$ 时，它等于 $7 - (0 + 1 + \cdots + (n-4))$ $= 7 - \frac{(n-4)(n-3)}{2}$ 代入 $n = 12$ ，得到 $7 - \frac{8 \cdot 9}{2} = -29$ 因此正确答案是 A。

Each interior entry feeds two entries below, and the end values $-1$ and $1$ cancel in the sum. So every row's total doubles the one above. The top row sums to $3,$ and $12{,}288 = 3 \cdot 2^{12},$ so this is the $12$ th row (counting the top as row $0$ ). Track the diagonals from the left. The second diagonal is $3, 2, 1, 0, -1, \ldots,$ dropping by $1$ each row. The third diagonal adds these up: for $n \ge 4$ it equals $7 - (0 + 1 + \cdots + (n-4))$ $= 7 - \frac{(n-4)(n-3)}{2}.$ Plug in $n = 12$ : $7 - \frac{8 \cdot 9}{2} = -29.$ Thus, A is the correct answer.

2025 AMC 10A 第 19 题

答案：A

难度评级：1910

解答：

其他年份的第 19 题