Find the sum of even numbers up to a number x javascript

Question:

I need to make a function that takes a number X as a parameter and that returns the total of the sum of all the numbers that are even from 0 to X.

This makes 0+2+4+6+8+10+…..X.

But I don't know what else I need to add, because this is what I have so far and it doesn't return what I ask for:

function sumaDeLosParesDel0Al(x){
      var suma = 0
      for (var i=1; i<x; i++){
        if (i%2 == 0) {
          suma += i
        }
      }
      return suma
    }
    
console.log(sumaDeLosParesDel0Al(4));

Answer:

I propose you this algorithm that does not use loops.

There are exactly N/2 number of even numbers between 2 and N , when N is even, and (N-1)/2 even numbers when N is odd.

The sum of the smallest ( N_0 ) of these even numbers with the largest of them ( N_n ), is equal to the sum of the next largest even number ( N_1 ) with the even number before the largest ( N_n-1 ).

This is an arithmetic progression , which translates into the following formula:

const suma = n*(min + max)/2;

Where n is the number of pairs between 0 and N , in this case min would be 2 and max will be equal to N when it is even and N-1 when it is odd.

So the algorithm can be as follows:

const sumaSoloPares = (numero) => {
  if (numero < 2) return 0;
  if (numero < 4) return 2;
  const numOfPares = Math.floor(numero/2);
  const min = 2;
  const max = numero%2 === 0 ? numero : numero - 1;
  const suma = ((min + max)*numOfPares)/2;
  return suma;
}

console.log(sumaSoloPares(10));
console.log(sumaSoloPares(20));
console.log(sumaSoloPares(1345));
console.log(sumaSoloPares(15));
console.log(sumaSoloPares(40));
console.log(sumaSoloPares(41));

I hope this provides another insight into solving the problem.