I've been working on the curriedSum exercise from intro-js/arguments.md. I've got the following recursive solution working, but not sure how the iterative version is supposed to work.

currySum

function currySumRecursive(n, sum) {
  //sum will not be passed in on the first call
	// so we set it to 0.
	if(sum === undefined) {
		sum = 0;
	}
	
	//if this is the last recursive call
	// return a simple function that returns 
	// the sum
	if(n===1) {
		return function(x) {
			return sum + x;
		}
	} 
	//otherwise recursively call currySumRecursive
	// passing the sum along until there are 
	// n curried (correct usage of term curried?) functions. 
	else {
		return function(x) {
			return currySumRecursive(n-1, x + sum);
		}
	}
}

var sumRecursive = currySumRecursive(1);
console.log("calling recursive sum (3): " + sumRecursive(3));

var sumRecursive2 = currySumRecursive(3);
console.log("calling recursive sum (1)(2)(3): " + sumRecursive2(1)(2)(3));

function currySumIterative(n) {
	var s = function(nums) {
		var sum = 0;
		nums.forEach(function(num) {
			sum += num;
		});
		return sum;
	}
	
	var args = undefined;
	
	var doll = function(x) {
		if(args === undefined) {
		  args = [].slice.call(arguments, 0);			
		}
		//eventually we will get a function
		// that looks like sum(1,2,3,4)
		// if n was 4. then we would like to sum 
		// the arguments
		if(args.length === n) {
			return s.call(null, args);
		} else {
			//not sure what should go here
			// ideally we want to return fn { return fn {return fn}}
			// until we get args.len = n
			return function(y) {
				args.push(y);
				
				//??
				return doll(args);
			}
		}
	}
	return doll;
}


var sumIterative = currySumIterative(1);
console.log("calling iterative sum (2): " + sumIterative(2));

var sumIterative2 = currySumIterative(3);
console.log("calling iterative sum (1)(2)(3): " + sumIterative(1)(2)(3));