面试题 递归算法1+2+....+100求和

var sum2 = 0;
var sum3 = 0;

function calc2(num) {
    if (num > 0) { //递归：在函数体中自己调用自己,实现一个简易的循环
       sum2 += num;
       num--;
       calc2(num);
       //arguments.callee(num);
     }
      return sum2;
}

function calc3(num) {
     if (num < 11) { //递归：在函数体中自己调用自己,实现一个简易的循环
         sum2 += num;
         num++;
         calc3(num);
          //arguments.callee(num);
    }
       return sum2;
    }


//var result2 = calc2(10);
var result2 = calc3(1);
console.log(result2);

function num(n) {
    if (n == 1) return 1;
    return num(n - 1) + n;
}
num(100);