1*2+2*3+3*4+4*5+...+999*1000

设an=n*(n+1)
所以∑(i*(i+1))=∑(i*i)+∑i
i=1`````999
分组求和得
1*2+2*3+3*4+4*5+...+999*1000
=999*(999+1)*(2*999+1)/6+(1+999)*999/2
=332833500+499500
=333333000

Pascal语言
var i:longint;
s:int64;
begin
s:=0;
for i:=1 to 999 do
s:=s+i*(i+1);
writeln(s);
readln
end.

1*2+2*3+3*4+...+999*1000
=999*(999+1)*(2*999+1)/6+(1+999)*999/2
=332833500+499500
=333333000

C++写的程序源码
#include <iostream>
using namespace std;

int main()
{
unsigned int sum=0;
for (int i=1; i<1000; i++)
{
sum += i * (i+1);

}
cout<<sum<<endl;
return 0;
}