Question

一个类似帕斯卡三角形的三角形，第一行为 $10$ ，第二行为 $10$ 后接 $1$ 。从之后的每一行开始，第一个数是 $10$ ，最后一个数是 $1$ ，并且和标准帕斯卡三角形一样，行中每个其他数都是它正上方两个数之和。前四行如下图所示。

第 $11$ 行所有数之和的各位数字之和是多少？

A Pascal-like triangle has $10$ as the top row and $10$ followed by $1$ as the second row. In each subsequent row the first number is $10,$ the last number is $1,$ and, as in the standard Pascal's Triangle, each other number in the row is the sum of the two numbers directly above it. The first four rows are shown below.

What is the sum of the digits of the sum of the numbers in the $11$ th row?

$11$

$13$

$14$

$16$

$17$

答案：D

知识点：杨辉三角递推等比数列

难度评级：1270

解答：

设 $S_n$ 为第 $n$ 行的和。第 $n-1$ 行的每个数都会贡献到下一行的两个位置，而固定的边界数 $10$ 和 $1$ 正好补上边缘缺少的项。因此 $S_n = 2 S_{n-1}$ （当 $n \ge 3$ ）。由 $S_2 = 11$ ，得到 $S_n = 11 \cdot 2^{n-2}$ ，所以 $S_{11} = 11 \cdot 2^9 = 5632$ 。其各位数字之和为 $5 + 6 + 3 + 2 = 16$ 。因此正确答案是 D。

Let $S_n$ be the sum of row $n.$ Each entry in row $n-1$ feeds the two entries just below it, and the fixed border numbers $10$ and $1$ exactly make up for the terms lost at the edges. So $S_n = 2 S_{n-1}$ for $n \ge 3.$ With $S_2 = 11,$ this gives $S_n = 11 \cdot 2^{n-2},$ so $S_{11} = 11 \cdot 2^9 = 5632.$ Its digits sum to $5 + 6 + 3 + 2 = 16.$ Thus, D is the correct answer.

2025 AMC 10B 第 3 题

答案：D

难度评级：1270

解答：

其他年份的第 3 题