Program for Sum of squares of first n natural numbers

Last Updated : 29 Mar, 2025

Given a positive integer n, we have to find the sum of squares of first n natural numbers.
Examples :

Input : n = 2
Output: 5
Explanation: 1^2+2^2 = 5

Input : n = 8
Output: 204
Explanation : 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 = 204

[Naive Approach] – Adding One By One – O(n) Time and O(1) Space

The idea for this naive approach is to run a loop from 1 to n and sum up all the squares.

C++

#include <iostream> using namespace std;  int summation(int n) {     int sum = 0;     for (int i = 1; i <= n; i++)         sum += (i * i);      return sum; }  int main() {     int n = 2;     cout << summation(n);     return 0; }

#include <stdio.h>   int summation(int n) {     int sum = 0;     for (int i = 1; i <= n; i++)         sum += (i * i);     return sum; }  int main() {     int n = 2;     printf("%d",summation(n));     return 0; }

Java

import java.util.*; import java.lang.*;  class GfG  {      public static int summation(int n)     {         int sum = 0;         for (int i = 1; i <= n; i++)             sum += (i * i);          return sum;     }       public static void main(String args[])     {         int n = 2;         System.out.println(summation(n));     } }

Python

def summation(n):     return sum([i**2 for i in                 range(1, n + 1)])  if __name__ == "__main__":     n = 2     print(summation(n))

using System; class GfG  {      public static int summation(int n)     {         int sum = 0;         for (int i = 1; i <= n; i++)             sum += (i * i);          return sum;     }      public static void Main()     {         int n = 2;          Console.WriteLine(summation(n));     } }

JavaScript

function summation(n) {     let sum = 0;     for (let i = 1; i <= n; i++)         sum += (i * i);      return sum; }   let n = 2; console.log(summation(n));

Output

[Expected Approach]- Using Mathematical Formulae – O(1) Time and O(1) Space

The idea for this approach is to use the mathematical formulae for the sum of squares of first n natural numbers.

1² + 2² + ……… + n² = n(n+1)(2n+1) / 6

We can prove this formula using induction. We can easily see that the formula is true for n = 1 and n = 2 as sums are 1 and 5 respectively.

Let it be true for n = k-1. So sum of k-1 numbers
is (k – 1) * k * (2 * k – 1)) / 6

In the following steps, we show that it is true
for k assuming that it is true for k-1.

Sum of k numbers = Sum of k-1 numbers + k²
= (k – 1) * k * (2 * k – 1) / 6 + k²
= ((k² – k) * (2*k – 1) + 6k²)/6
= (2k³ – 2k² – k² + k + 6k²)/6
= (2k³ + 3k² + k)/6
= k * (k + 1) * (2*k + 1) / 6

Example : Find sum of squares of the first 3 natural numbers
Solution:
= 3 * (3 + 1) * (2*3 + 1) / 6
= (3 * 4 * 7) / 6
= 84 / 6
= 14

C++

#include <iostream> using namespace std;   int summation(int n) {     return (n * (n + 1) *          (2 * n + 1)) / 6; }  int main() {     int n = 10;     cout << summation(n) << endl;     return 0; }

#include <stdio.h>   int summation(int n) {     return (n * (n + 1) *             (2 * n + 1)) / 6; }  int main() {     int n = 10;     printf("%d", summation(n));     return 0; }

Java

import java.util.*; import java.lang.*;  class GFG  {      public static int summation(int n)     {         return (n * (n + 1) *                 (2 * n + 1)) / 6;     }       public static void main(String args[])     {         int n = 10;         System.out.println(summation(n));     } }

Python

# Python code to find sum of  # squares of first n natural numbers. def summation(n):     return (n * (n + 1) *             (2 * n + 1)) / 6      # Driver Code if __name__ == '__main__':     n = 10     print(summation(n))

using System;  class GFG  {       public static int summation(int n)     {         return (n * (n + 1) *                 (2 * n + 1)) / 6;     }       public static void Main()     {         int n = 10;          Console.WriteLine(summation(n));     } }

JavaScript

function summation(n) { return (n * (n + 1) * (2 * n + 1)) / 6; } let n = 10; console.log(summation(n));

Output

Avoiding the overflow:
In the above method, sometimes due to large value of n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid this overflow up to some extent using the fact that n*(n+1) must be divisible by 2 and restructuring the formula as (n * (n + 1) / 2) * (2 * n + 1) / 3;

C++

#include <iostream> using namespace std;   int summation(int n) { //to avoid overflow  //n*(n + 1)*(2 * n + 1) / 6 = (n * (n + 1) / 2) * (2 * n + 1) / 3;     return (n * (n + 1) / 2) * (2 * n + 1) / 3; }   int main() {     int n = 10;     cout << summation(n) << endl;     return 0; }

Java

import java.io.*;  class GFG {     static int summation(int n)   {     //to avoid overflow      //n*(n + 1)*(2 * n + 1) / 6 = (n * (n + 1) / 2) * (2 * n + 1) / 3;     return (n * (n + 1) / 2) * (2 * n + 1) / 3;   }    public static void main (String[] args) {     int n = 10;     System.out.println(summation(n));    } }

Python

def summation(n):     #to avoid overflow      #n*(n + 1)*(2 * n + 1) / 6 = (n * (n + 1) / 2) * (2 * n + 1) / 3;     return (n * (n + 1) // 2) * (2 * n + 1) // 3  def main():     n = 10     print(summation(n))  if __name__ == "__main__":     main()

using System; class MainClass  {    public static int Summation(int n)   {     //to avoid overflow      //n*(n + 1)*(2 * n + 1) / 6 = (n * (n + 1) / 2) * (2 * n + 1) / 3;     return (n * (n + 1) / 2) * (2 * n + 1) / 3;   }    public static void Main(string[] args)   {     int n = 10;     Console.WriteLine(Summation(n));   } }

JavaScript

function summation(n) {     //to avoid overflow      //n*(n + 1)*(2 * n + 1) / 6 = (n * (n + 1) / 2) * (2 * n + 1) / 3;     return (n * (n + 1) / 2) * (2 * n + 1) / 3; }  n = 10; console.log(summation(n));

Output

Sum of squares of first n natural numbers

Rahul_ and KANCHAN RAY

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Program for Sum of squares of first n natural numbers

[Naive Approach] – Adding One By One – O(n) Time and O(1) Space

[Expected Approach]- Using Mathematical Formulae – O(1) Time and O(1) Space

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