Sum of Squares

Easy Loops
Find the sum of squares of the first n natural numbers using a loop.

Sample input

5

Sample output

55

Solution

n = 5
print(sum(i * i for i in range(1, n + 1)))
const n = 5;
let total = 0;
for (let i = 1; i <= n; i++) total += i * i;
console.log(total);
public class Main {
    public static void main(String[] args) {
        int n = 5, total = 0;
        for (int i = 1; i <= n; i++) total += i * i;
        System.out.println(total);
    }
}
fun main() {
    val n = 5
    val total = (1..n).sumOf { it * it }
    println(total)
}
let n = 5
let total = (1...n).reduce(0) { $0 + $1 * $1 }
print(total)
void main() {
  int n = 5, total = 0;
  for (var i = 1; i <= n; i++) total += i * i;
  print(total);
}
#include <iostream>
using namespace std;

int main() {
    int n = 5, total = 0;
    for (int i = 1; i <= n; i++) total += i * i;
    cout << total << endl;
    return 0;
}
#include <stdio.h>

int main() {
    int n = 5, total = 0;
    for (int i = 1; i <= n; i++) total += i * i;
    printf("%d\n", total);
    return 0;
}