Simpler solution of the iterative method

multiples.js

function sumMultiples(n, mul){
  var sum=0;
  for (var i = 0 ; i < n ; i = mul + i) {
    sum += i;
  }
  return sum;
}

console.log(sumMultiples(1000, 5) + sumMultiples(1000, 3) - sumMultiples(1000, 15));