从1到n这n个连续自然数的求和公式为1+2+3+…+n=n(n+1)/2.灵活巧用它,一些问题的解答可变得简易、迅捷.现举例如下:一、有理数计算问题例1计算1+2+…+98+99+98+…+2+1/99×99.解析:分子是197个数相加,其中前98个加数是从1开始顺数加到98,后98个加数是从98开始倒数加数1,那么 The summation formula of n consecutive natural numbers from 1 to n is 1 + 2 + 3 + ... + n = n (n + 1) / 2. It is flexible to use it, and the solution of some problems can be simplified and quick. Examples are as follows: First, the rational number calculation problem Example 1 Calculate 1 + 2 + ... + 98 + 99 + 98 + ... + 2 + 1/99 × 99. Analysis: The numerator is 197 numbers added, of which the first 98 plus Starting from 1, the number is added to 98, the last 98 plus is the countdown 1 from 98, then