What’s your alternative solution? Challenge #5

About this series

This is series of daily JavaScript coding challenges... for both beginners and advanced users.

Each day I’m gone present you a very simple coding challenge, together with the solution. The solution is intentionally written in a didactic way using classic JavaScript syntax in order to be accessible to coders of all levels.

Solutions are designed with increase level of complexity.

Today’s coding challenge

Calculate the sum of numbers from 1 to 10.

(scroll down for solution)

Code newbies

If you are a code newbie, try to work on the solution on your own. After you finish it, or if you need help, please consult the provided solution.

Advanced developers

Please provide alternative solutions in the comments below.

You can solve it using functional concepts or solve it using a different algorithm... or just solve it using the latest ES innovations.

By providing a new solution you can show code newbies different ways to solve the same problem.

Solution

// Solution for challenge05

var sum = 0;

for(var i = 1; i <= 10; i++)
{
    sum += i;
}

println(sum);

To quickly verify this solution, copy the code above in this coding editor and press "Run".

Note: The solution was originally designed for codeguppy.com environment, and therefore is making use of println. This is the almost equivalent of console.log in other environments. Please feel free to use your preferred coding playground / environment when implementing your solution.

Comments

[Shankar Sridhar]

const sumFrom1ToN = n => (n * (n + 1)) / 2;

This will work with any number of input numbers, as long as input starts from 1 to n and they are all integers.

Sample Runs:
sumFrom1ToN(10); // 55
sumFrom1ToN(5); // 15

[Shankar Sridhar]

var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
numbers.reduce((acc, item) => acc + number, 0);

[George WL]

I'd suggest using CodePen, it's got a far better interface than codeguppy

[George WL]

Could be be adapted to a oneliner, but I prefer readability over brevity.

const numbers = [...Array(10)].map((_,i)=>i+1);
const solution = [...numbers].reduce((a,b)=>a+b);

Adapted to a function which can take any number higher than 0:

function getSumOneToN(n){
     const numbers = [...Array(n)].map((_,i)=>i+1);
     return [...numbers].reduce((a,b)=>a+b);
}