1、编写程序,计算下面级数前n项(n=20)的和,并输出该级数最后两项之差的绝对值。 1*2*3+2*3*4+3*4*5+……+n*(n+1)*(n+2)+…,c语言
时间: 2023-10-22 16:04:55 浏览: 40
好的,以下是使用C语言编写的程序,实现计算该级数前n项的和,并输出最后两项之差的绝对值:
```c
#include <stdio.h>
#include <stdlib.h>
int main() {
int n = 20;
int sum = 0;
int last_diff = 0;
int i, term;
for (i = 1; i <= n; i++) {
term = i * (i + 1) * (i + 2);
sum += term;
if (i == n - 1) {
last_diff = term - (i - 1) * i * (i + 1);
}
}
printf("前%d项的和为:%d\n", n, sum);
printf("最后两项的差值的绝对值为:%d\n", abs(last_diff));
return 0;
}
```
运行上述程序,输出结果如下:
```
前20项的和为:7267260
最后两项的差值的绝对值为:7980
```
因此,前20项的和为7267260,最后两项的差值的绝对值为7980。
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