Sum of square-sums of first n natural numbers

Namachila November 19, 2025

0 0 6 minutes read

Given a positive integer n. The task is to find the sum of the sum of square of first n natural number.
Examples :

Input : n = 3
Output : 20
Sum of square of first natural number = 1
Sum of square of first two natural number = 1^2 + 2^2 = 5
Sum of square of first three natural number = 1^2 + 2^2 + 3^2 = 14
Sum of sum of square of first three natural number = 1 + 5 + 14 = 20

Input : n = 2
Output : 6

Method 1: O(n) The idea is to find sum of square of first i natural number, where 1 <= i <= n. And then add them all.
We can find sum of squares of first n natural number by formula: n * (n + 1)* (2*n + 1)/6
Below is the implementation of this approach:

C++

// CPP Program to find the sum of sum of 
// squares of first n natural number
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of sum of square of 
// first n natural number
int findSum(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; i++) 
        sum += ((i * (i + 1) * (2 * i + 1)) / 6);
    return sum;
}
 
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java Program to find the sum of 
// sum of squares of first n natural
// number
 
class GFG {
     
    // Function to find sum of sum of
    // square of first n natural number
    static int findSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++) 
            sum += ((i * (i + 1) 
               * (2 * i + 1)) / 6);
        return sum;
    }
     
    // Driver Program
    public static void main(String[] args)
    {
        int n = 3;
     
        System.out.println( findSum(n));
    }
}
 
 
// This code is contributed by 
// Arnab Kundu

Python3

# Python3 Program to find the sum
# of sum of squares of first n 
# natural number
 
# Function to find sum of sum of
# square of first n natural number
def findSum(n):
    summ = 0
    for i in range(1, n+1):
        summ = (summ + ((i * (i + 1) 
                * (2 * i + 1)) / 6)) 
    return summ 
 
# Driven Program
n = 3
print(int(findSum(n)))
 
 
# This code is contributed by 
# Prasad Kshirsagar

C#

// C# Program to find the sum of sum of 
// squares of first n natural number
using System;
 
public class GFG {
     
    // Function to find sum of sum of
    // square of first n natural number
    static int findSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++) 
            sum += ((i * (i + 1) * 
                  (2 * i + 1)) / 6);
        return sum;
    }
     
    // Driver Program
    static public void Main()
    {
        int n = 3;
         
        Console.WriteLine(findSum(n));
    }
}
 
// This code is contributed by
// Arnab Kundu.

PHP

<?php
// PHP Program to find the sum of 
// squares of first n natural number
 
// Function to find sum of sum of 
// square of first n natural number
function findSum( $n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++) 
        $sum += (($i * ($i + 1) * 
                 (2 * $i + 1)) / 6);
    return $sum;
}
 
// Driver Code
$n = 3;
echo findSum($n) ;
 
// This code is contributed by anuj_67.
?>

Javascript

<script>
 
// Javascript Program to find the sum of sum of 
// squares of first n natural number 
 
// Function to find sum of sum of square 
// of first n natural number 
function findSum(n) 
{ 
    return (n * (n + 1) * (n + 1) * (n + 2)) / 12; 
} 
 
// Driven Program 
 
    let n = 3; 
    document.write(findSum(n) + "<br>"); 
 
 
// This code is contributed by Mayank Tyagi
 
</script>

Output :

Time Complexity : O(n)
Auxiliary Space : O(1)
Method 2: O(1)
Mathematically, we need to find, ? ((i * (i + 1) * (2*i + 1)/6), where 1 <= i <= n
So, lets solve this summation,

Sum = ? ((i * (i + 1) * (2*i + 1)/6), where 1 <= i <= n
    = (1/6)*(? ((i * (i + 1) * (2*i + 1)))
    = (1/6)*(? ((i² + i) * (2*i + 1))
    = (1/6)*(? ((2*i³ + 3*i² + i))
    = (1/6)*(? 2*i³ + ? 3*i² + ? i)
    = (1/6)*(2*? i³ + 3*? i² + ? i)
    = (1/6)*(2*(i*(i + 1)/2)² + 3*(i * (i + 1) * (2*i + 1))/6 + i * (i + 1)/2)
    = (1/6)*(i * (i + 1))((i*(i + 1)/2) + ((2*i + 1)/2) + 1/2)
    = (1/12) * (i * (i + 1)) * ((i*(i + 1)) + (2*i + 1) + 1)
    = (1/12) * (i * (i + 1)) * ((i² + 3 * i + 2)
    = (1/12) * (i * (i + 1)) * ((i + 1) * (i + 2))
    = (1/12) * (i * (i + 1)² * (i + 2))

Below is the implementation of this approach:

C++

// CPP Program to find the sum of sum of 
// squares of first n natural number
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of sum of square
// of first n natural number
int findSum(int n)
{
    return (n * (n + 1) * (n + 1) * (n + 2)) / 12;
}
 
// Driven Program
int main()
{
    int n = 3;
    cout << findSum(n) << endl;
    return 0;
}

Java

// Java Program to find the sum of sum of 
// squares of first n natural number
class GFG {
 
    // Function to find sum of sum of 
    // square of first n natural number
    static int findSum(int n)
    {
        return (n * (n + 1) *
        (n + 1) * (n + 2)) / 12;
    }
     
    // Driver Program
    public static void main(String[] args)
    {
        int n = 3;
         
        System.out.println(findSum(n) );
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 Program to find the sum 
# of sum of squares of first n 
# natural number
 
# Function to find sum of sum of
# square of first n natural number
def findSum(n):
    return ((n * (n + 1) * (n + 1)
                 * (n + 2)) / 12)
 
# Driven Program
n = 3
print(int(findSum(n)))
 
# This code is contributed by 
# Prasad Kshirsagar

C#

// C# Program to find the sum of sum of 
// squares of first n natural number
using System;
 
class GFG {
 
    // Function to find sum of sum of
    // square of first n natural number
    static int findSum(int n)
    {
        return (n * (n + 1) * (n + 1)
                     * (n + 2)) / 12;
    }
     
    // Driver Program
    static public void Main()
    {
        int n = 3;
         
        Console.WriteLine(findSum(n) );
    }
}
 
// This code is contributed by Arnab Kundu

PHP

<?php
// PHP Program to find the sum of sum of 
// squares of first n natural number
 
// Function to find sum of sum of square
// of first n natural number
function findSum($n)
{
    return ($n * ($n + 1) * 
           ($n + 1) * ($n + 
                   2)) / 12;
}
 
    // Driver Code
    $n = 3;
    echo findSum($n) ;
 
// This code is contributed by nitin mittal
?>

Javascript

<script>
// js Program to find the sum of sum of
// squares of first n natural number
 
// Function to find sum of sum of square
// of first n natural number
function findSum($n)
{
    return (n * (n + 1) *(n + 1) * (n + 2)) / 12;
}
 
    // Driver Code
    n = 3;
    document.write(findSum(n)) ;
 
// This code is contributed by sravan kumar
</script>

Output:

Time Complexity : O(1)
Auxiliary Space : O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!