AIME 1985 · 第 5 题

Solution

The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$, and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$.

Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \cdot 248$ is the largest multiple of 6 less than 1492, so

$$\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}$$ $$=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,$$ where we can make this last step because $\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero.

Similarly, since $1980 = 6 \cdot 330$, $\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$.

Finally, $\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$.

Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\cdot 493 = \boxed{986}$.