求1000以内的奇数和

(1+999)*500/2

等差数列前n项和, 一共500项
S(n) = n*n
#include <stdio.h>
unsigned long sum(unsigned long);
int main (void) {
printf("%ld\n", sum(500));
return 0;
}

unsigned long sum (unsigned long n) {
return n*n;
}

@echo off
for /l %%i in (1,2,999) do set /a n+=%%i
echo %n%
pause
存为a.bat文件。

var
a,s:integer;
begin
s:=0;
for a:=1 to 1000 do
begin
if a mod 2=1 then s:=s+a;
end;
writeln(s);
end.

int sum = 0;
for(int i = 1; i < 1000; i += 2)
{
sum += i;
}
printf("%d", sum);