Question

用C语言解决1*2*3+3*4*5+5*6*7+……99*100*101（具体程序）

Answer 1

static void Main(string[] args)
{
int a = 0, b = 1, c = 2, d = 3;//结果a，初始值b，c，d
int e,f,g,h;//过程值e，f，g，h
while (b<=99)
{
e = b * c * d;
f=b+2;
g=c+2;
h = d + 2;
b = f;
c = g;
d = h;
a+=e;
}
Console.WriteLine("1*2*3+3*4*5+5*6*7+……+99*100*101的结果为" + a);
Console.ReadKey();

Answer 2

#include <stdio.h>
int main()
{
int i;
long sum=0;
for(i=1;i<=99;i+=2) sum+=i*(i+1)*(i+2);
printf("The result is %ld\n",sum);
return 0;
}

Answer 3

#include <stdio.h>
void main()
{
long s=0;
int i;
for(i=3;i<=101;i+=2)
{
s+=(i-1)*(i-2)*i;
printf("%d*%d*%d",(i-2),(i-1),i);
if(i==101)printf("=%d\n",s);
else printf("+");
}
}