1985 AIME 第 5 题

Question

一个整数数列

a_1

、

a_2

、

a_3

、

\ldots

满足：对每个

n\geq3

，都有

a_n=a_{n-1}-a_{n-2}

。若该数列前

1492

项的和为

1985

，且前

1985

项的和为

1492

，求前

2001

项的和。

A sequence of integers

a_1,

a_2,

a_3,

\ldots

is chosen so that

a_n=a_{n-1}-a_{n-2}

for each

n\geq3.

What is the sum of the first

2001

terms of this sequence if the sum of the first

1492

terms is

1985,

and the sum of the first

1985

terms is

1492?

答案

答案：986

知识点：递推找规律方程组

难度评级：2110

小提示：

用

a_1

和

a_2

表示前六项

Write the first six terms in terms of

a_1

and

a_2

大提示：

该数列每六项循环一次，而且每个六项组的和为

0

The sequence repeats every six terms, and each six-term block has sum

0

解答：

令

a_1=x

，且

a_2=y

。前六项为

x,\ y,\ y-x,\ -x,\ -y,\ x-y\text{，}

此后数列开始重复；这六项之和为

0

。由于

1492\equiv4\pmod6

，且

1985\equiv5\pmod6

，已知条件给出

\begin{aligned} 2y-x&=1985,\\ y-x&=1492 \end{aligned}\text{。}

因而

y=493

。又因为

2001\equiv3\pmod6

，所求和为

x+y+(y-x)=2y=986

。

Put

a_1=x

and

a_2=y.

The first six terms are

x,\ y,\ y-x,\ -x,\ -y,\ x-y,

after which the sequence repeats; these six terms sum to

0.

Since

1492\equiv4\pmod6

and

1985\equiv5\pmod6,

the given equations are

\begin{aligned} 2y-x&=1985,\\ y-x&=1492. \end{aligned}

Thus

y=493.