const sumAll = ([x, y]) => (x + y) * (Math.abs(x - y) + 1) / 2;

// const sum_range = (x, y) => (x + y) * (abs(x - y) + 1) / 2;

// const sumAll = (arr) => (Math.max(...arr)*(Math.max(...arr)+1) / 2) - (Math.min(...arr)*(Math.min(...arr)-1) / 2);

// This is a better solution using actual Math.

// Example 1-10 adds up to 10 * (10 + 1) / 2 = 55

/*

function sumAll(arr) {

function sumRange(a, b) {

return (a*(a+1) / 2) - (b*(b-1) / 2);

}

return sumRange(Math.max(...arr), Math.min(...arr));

}

*/

// const sumAll = (arr) => arr.sort((a, b) => b - a).reduce((pr, cu) => (pr*(pr+1)/2) - (cu*(cu-1)/2));

/*

With this solution, I was playing around with creating a long array.

I grabbed some of the code off of Array.from() and modified it.

const sumAll = (arr) => Array.from({length: Math.max(...arr) - Math.min(...arr) + 1}, (v, k) => k + Math.min(...arr)).reduce((pr, cu) => pr + cu);

*/

/*

function sumAll(arr) {

arr = arr.sort(function(a,b) {

return a - b;

});

var sum = 0;

for (var i = arr[0]; i <= arr[arr.length-1]; i++) {

sum += i;

}

return sum;

}

*/

sumAll([1, 4]); // should return a number.

sumAll([1, 4]); // should return 10.

sumAll([4, 1]); // should return 10.

sumAll([5, 10]); // should return 45.

sumAll([10, 5]); // should return 45.