1992 AIME 第 4 题

Question

在帕斯卡三角形中，每个数都是它上方两个数之和。该三角形的前几行如下。

\begin{array}{rl}0:&1\\1:&1\quad1\\2:&1\quad2\quad1\\3:&1\quad3\quad3\quad1\\4:&1\quad4\quad6\quad4\quad1\\5:&1\quad5\quad10\quad10\quad5\quad1\\6:&1\quad6\quad15\quad20\quad15\quad6\quad1\end{array}

帕斯卡三角形的哪一行中有三个连续的数，其比为

3:4:5

？

In Pascal’s Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.

\begin{array}{rl}0:&1\\1:&1\quad1\\2:&1\quad2\quad1\\3:&1\quad3\quad3\quad1\\4:&1\quad4\quad6\quad4\quad1\\5:&1\quad5\quad10\quad10\quad5\quad1\\6:&1\quad6\quad15\quad20\quad15\quad6\quad1\end{array}

In which row of Pascal’s Triangle do three consecutive entries occur that are in the ratio

3:4:5?

答案

将这三个数表示为

\binom nk

、

\binom n{k+1}

和

\binom n{k+2}

Represent the three entries as

\binom nk,

\binom n{k+1},

and

\binom n{k+2}

利用相邻二项式系数之比，得到关于

n

和

k

的两个一次方程

Use the ratios of consecutive binomial coefficients to obtain two linear equations in

n

and

k

对从位置

k

开始的三个连续数，有

\begin{aligned}\frac{\binom n{k+1}}{\binom nk}&=\frac{n-k}{k+1}=\frac43,\\\frac{\binom n{k+2}}{\binom n{k+1}}&=\frac{n-k-1}{k+2}=\frac54\end{aligned}\text{。}

因此

3n=7k+4

，且

4n=9k+14

。解得

k=26

、

n=62

。

For three consecutive entries beginning at position

k,

\begin{aligned}\frac{\binom n{k+1}}{\binom nk}&=\frac{n-k}{k+1}=\frac43,\\\frac{\binom n{k+2}}{\binom n{k+1}}&=\frac{n-k-1}{k+2}=\frac54.\end{aligned}

Thus

3n=7k+4

and

4n=9k+14.

Solving gives

k=26

and

n=62.

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