Sum of square-sums of first n natural numbers

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 them add them all.
We can find sum of squares of first n natural number by fromula: n * (n + 1)* (2*n + 1)/6

Below is the implementation of this approach:

C++



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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

20

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

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

Sum = Σ ((i * (i + 1) * (2*i + 1)/6), where 1 <= i <= n
    = (1/6)*(Σ ((i * (i + 1) * (2*i + 1)))
    = (1/6)*(Σ ((i2 + i) * (2*i + 1))
    = (1/6)*(Σ ((2*i3 + 3*i2 + i))
    = (1/6)*(Σ 2*i3 + Σ 3*i2 + Σ i)
    = (1/6)*(2*Σ i3 + 3*Σ i2 + Σ 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 implentation of this approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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code

<?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
?>

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Output:

20

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



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