Sum of Natural Numbers

Easy Numbers

Find the sum of the first n natural numbers (1 + 2 + ... + n).

Solution

n = 10
print(n * (n + 1) // 2)

const n = 10;
console.log((n * (n + 1)) / 2);

public class Main {
    public static void main(String[] args) {
        int n = 10;
        System.out.println(n * (n + 1) / 2);
    }
}

fun main() {
    val n = 10
    println(n * (n + 1) / 2)
}

let n = 10
print(n * (n + 1) / 2)

void main() {
  int n = 10;
  print(n * (n + 1) ~/ 2);
}

#include <iostream>
using namespace std;

int main() {
    int n = 10;
    cout << n * (n + 1) / 2 << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int n = 10;
    printf("%d\n", n * (n + 1) / 2);
    return 0;
}