AMC10 2025 A

AMC10 2025 A · Q19

AMC10 2025 A · Q19. It mainly tests Polynomials, Sequences & recursion (algebra).

An array of numbers is constructed beginning with the numbers $-1$, $3$, and $1$ in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with $-1$ and $1,$ respectively. \[\large{-1}\qquad\large{3}\qquad\large{1}\] \[\large{-1}\qquad\large{2}\qquad\large{4}\qquad\large{1}\] \[\large{-1}\qquad\large{1}\qquad\large{6}\qquad\large{5}\qquad\large{1}\] If the process continues, one of the rows will sum to $12{,}288$. In that row, what is the third number from the left?

一个数字阵列从顶行数字 $-1$、$3$ 和 $1$ 开始构造。每相邻一对数字相加产生下一行的数字。每行开始和结束分别为 $-1$ 和 $1$。 \[\large{-1}\qquad\large{3}\qquad\large{1}\] \[\large{-1}\qquad\large{2}\qquad\large{4}\qquad\large{1}\] \[\large{-1}\qquad\large{1}\qquad\large{6}\qquad\large{5}\qquad\large{1}\] 如果过程继续，有一行之和为 $12{,}288$。在那一行中，距左边第三个数是多少？

(A) \: -29 \: -29

(B) \: -21 \: -21

(D) \: -18 \: -18

(E) \: -3 \: -3

Answer

Correct choice: (A)

正确答案：（A）

Solution

Consider the polynomial $f(x) = -x^2+3x+1.$ When we multiply this polynomial by $x+1,$ we are essentially doing the operation given in the problem (When we multiply $p(x)$ by $x+1,$ a term of degree $d$ in the yielded expression is the sum of $1\cdot(\text{degree d})$ and $x\cdot(\text{degree d-1})$ in $p(x)$ This effect is visible in Pascal's Triangle). So, if we let the coefficients of $f(x)$ be the zero row of the array, then the $n^{th}$ row is just the coefficients of $f(x)(x+1)^n.$ The next thing to note is that the sum of the coefficients in any polynomial $p(x)$ is just $p(1).$ Therefore, the sum of the entries in the $n^{th}$ row of the array is $f(1)(1+1)^n=3\cdot2^n.$ Letting this equal $12288,$ we get $n=12.$ We are looking for the $3^{rd}$ term in the $12^{th}$ row. The $12^{th}$ row is given by the coefficients of $f(x)(x+1)^{12}=(-x^2+3x+1)(x+1)^{12}.$ Since the degree of the resulting expression is $14,$ the third term in the row is just the coefficient of $x^{12}$ in the expression, which is $-\dbinom{12}{10}+3\dbinom{12}{11}+1=\boxed{\textbf{(A) }-29}.$

考虑多项式 $f(x) = -x^2+3x+1$。将此多项式乘以 $x+1$，本质上就是题目给出的操作（将 $p(x)$ 乘以 $x+1$，所得表达式中次数为 $d$ 的项是 $p(x)$ 中 $1\cdot$(次数 $d$) 和 $x\cdot$(次数 $d-1$) 的和，这种效果在帕斯卡三角中可见）。因此，如果让 $f(x)$ 的系数为阵列的零行，则第 $n$ 行就是 $f(x)(x+1)^n$ 的系数。接下来注意到，任意多项式 $p(x)$ 的系数和就是 $p(1)$。因此，阵列第 $n$ 行的条目和为 $f(1)(1+1)^n=3\cdot2^n$。设此等于 $12288$，得 $n=12$。我们要找第 $12$ 行的第 $3$ 项。第 $12$ 行由 $f(x)(x+1)^{12}=(-x^2+3x+1)(x+1)^{12}$ 的系数给出。结果表达式次数为 $14$，因此行中第三项就是表达式中 $x^{12}$ 的系数，为 $-\dbinom{12}{10}+3\dbinom{12}{11}+1=\boxed{\textbf{(A) }-29}$。

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