n elements of an array to create the product of those elements. Using a for loop, you could do this:
```js
  function multiply(arr, n) {
    var product = 1;
    for (var i = 0; i < n; i++) {
        product *= arr[i];
    }
    return product;
  }
```
However, notice that multiply(arr, n) == multiply(arr, n - 1) * arr[n - 1]. That means you can rewrite multiply in terms of itself and never need to use a loop.
```js
  function multiply(arr, n) {
    if (n <= 0) {
      return 1;
    } else {
      return multiply(arr, n - 1) * arr[n - 1];
    }
  }
```
The recursive version of multiply breaks down like this. In the base case, where n <= 0, it returns 1. For larger values of n, it calls itself, but with n - 1. That function call is evaluated in the same way, calling multiply again until n <= 0.  At this point, all the functions can return and the original multiply returns the answer.
Note: Recursive functions must have a base case when they return without calling the function again (in this example, when n <= 0), otherwise they can never finish executing.
sum(arr, n), that returns the sum of the first n elements of an array arr.
sum([1], 0) should equal 0.
    testString: assert.equal(sum([1], 0), 0);
  - text: sum([2, 3, 4], 1) should equal 2.
    testString: assert.equal(sum([2, 3, 4], 1), 2);
  - text: sum([2, 3, 4, 5], 3) should equal 9.
    testString: assert.equal(sum([2, 3, 4, 5], 3), 9);
  - text: Your code should not rely on any kind of loops (for or while or higher order functions such as forEach, map, filter, or reduce.).
    testString: assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
  - text: You should use recursion to solve this problem.
    testString: assert(removeJSComments(sum.toString()).match(/sum\(.*\)/g).length > 1);
```