Question

已知 $A_k = \frac{k(k - 1)}{2}\cos\frac{k(k - 1)\pi}{2}$ ，求 $|A_{19} + A_{20} + \cdots + A_{98}|$ 。

Given that $A_k = \frac{k(k - 1)}{2}\cos\frac{k(k - 1)\pi}{2},$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$

答案

答案：40

难度评级：2400

解答：

因为 $k(k-1)$ 为偶数， $n_k = \frac{k(k-1)}{2}$ 是整数，且 $\cos\frac{k(k-1)\pi}{2}$ $= \cos(n_k \pi)$ $= (-1)^{n_k}$ 。三角形数 $n_k$ 的奇偶性只取决于 $k \bmod 4$ ：当 $k \equiv 0, 1 \pmod 4$ 时为偶数，当 $k \equiv 2, 3 \pmod 4$ 时为奇数。所以当 $k \equiv 0, 1 \pmod 4$ 时 $A_k = n_k$ ，当 $k \equiv 2, 3 \pmod 4$ 时 $A_k = -n_k$ 。

从 $k = 19 \equiv 3 \pmod 4$ 开始，把这 $80$ 项分成 $20$ 个连续四项组。利用 $n_{j+1} - n_j = j$ ，每个 $k \equiv 3 \pmod 4$ 的组都化简为 $\begin{aligned} &A_k + A_{k+1} \\ &\quad {}+ A_{k+2} + A_{k+3} \\ &= (n_{k+1} - n_k) \\ &\quad {}- (n_{k+3} - n_{k+2}) \\ &= k - (k + 2) \\ &= -2. \end{aligned}$

总和为 $20 \cdot (-2) = -40$ ，所以所求绝对值为 $40$ 。

Since $k(k-1)$ is even, $n_k = \frac{k(k-1)}{2}$ is an integer and $\cos\frac{k(k-1)\pi}{2}$ $= \cos(n_k \pi)$ $= (-1)^{n_k}.$ The parity of the triangular number $n_k$ depends only on $k \bmod 4:$ it is even for $k \equiv 0, 1 \pmod 4$ and odd for $k \equiv 2, 3 \pmod 4.$ So $A_k = n_k$ when $k \equiv 0, 1 \pmod 4$ and $A_k = -n_k$ when $k \equiv 2, 3 \pmod 4.$

Group the $80$ terms into $20$ consecutive blocks of four starting at $k = 19 \equiv 3 \pmod 4.$ Using $n_{j+1} - n_j = j,$ each block with $k \equiv 3 \pmod 4$ collapses: $\begin{aligned} &A_k + A_{k+1} \\ &\quad {}+ A_{k+2} + A_{k+3} \\ &= (n_{k+1} - n_k) \\ &\quad {}- (n_{k+3} - n_{k+2}) \\ &= k - (k + 2) \\ &= -2. \end{aligned}$

The total is $20 \cdot (-2) = -40,$ so the requested absolute value is $40.$

1998 AIME 第 5 题

答案：40

难度评级：2400

解答：

其他年份的第 5 题