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Sum of cubes of first n odd natural numbers

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Given a number n, find sum of first n odd natural numbers.
 

Input  : 2
Output : 28
1^3 + 3^3 = 28

Input  : 4
Output : 496
1^3 + 3^3 + 5^3 + 7^3 = 496

 

A simple solution is to traverse through n odd numbers and find the sum of cubes. 
 

C++




// Simple C++ method to find sum of cubes of
// first n odd numbers.
#include <iostream>
using namespace std;
  
int cubeSum(int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += (2*i + 1)*(2*i + 1)*(2*i + 1);
    return sum;
}
  
int main()
{
    cout << cubeSum(2);
    return 0;
}


Java




// Java program to perform sum of
// cubes of first n odd natural numbers
  
public class GFG 
{
  
    public static int cubesum(int n)
    {
        int sum = 0;
        for(int i = 0; i < n; i++)
            sum += (2 * i + 1) * (2 * i +1
                   * (2 * i + 1);
                  
        return sum;
    }
      
  
    // Driver function
    public static void main(String args[])
    {
        int a = 5;
        System.out.println(cubesum(a));
          
    }
}
  
// This article is published Akansh Gupta


Python3




# Python3 program to find sum of 
# cubes of first n odd numbers.
  
def cubeSum(n):
    sum = 0
      
    for i in range(0, n) :
        sum += (2 * i + 1) * (2 * i + 1) * (2 * i + 1)
    return sum
  
# Driven code 
print(cubeSum(2))
  
# This code is contributed by Shariq Raza


C#




// C# program to perform sum of
// cubes of first n odd natural numbers
using System;
  
public class GFG 
{
  
    public static int cubesum(int n)
    {
        int sum = 0;
        for(int i = 0; i < n; i++)
            sum += (2 * i + 1) * (2 * i +1) 
                   * (2 * i + 1);
                  
        return sum;
    }
      
  
    // Driver function
    public static void Main()
    {
        int a = 5;
        Console.WriteLine(cubesum(a));
          
    }
}
  
// This code is published vt_m


PHP




<?php
// Simple PHP method to find sum of 
// cubes of first n odd numbers.
  
function cubeSum($n)
{
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += (2 * $i + 1) * 
                (2 * $i + 1) * 
                (2 * $i + 1);
    return $sum;
}
  
// Driver Code
echo cubeSum(2);
  
// This code is contributed by vt_m.
?>


Javascript




<script>
// Simple javascript method to find sum of cubes of
// first n odd numbers.
function cubeSum( n)
{
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += (2*i + 1)*(2*i + 1)*(2*i + 1);
    return sum;
}
  
    document.write(cubeSum(2));
  
// This code is contributed by Rajput-Ji
  
</script>


Output : 
 

28

Complexity Analysis:

Time Complexity: O(n), as we are using a single traversal in the cubeSum() function.

Space Complexity:O(1)

An efficient solution is to apply the below formula.
 

sum = n2(2n2 - 1) 

How does it work? 

We know that sum of cubes of first 
n natural numbers is = n2(n+1)2 / 4

Sum of first n even numbers is 2 *  n2(n+1)2 

Sum of cubes of first n odd natural numbers = 
            Sum of cubes of first 2n natural numbers - 
            Sum of cubes of first n even natural numbers 

         =  (2n)2(2n+1)2 / 4 - 2 *  n2(n+1)2 
         =  n2(2n+1)2 - 2 *  n2(n+1)2 
         =  n2[(2n+1)2 - 2*(n+1)2]
         =  n2(2n2 - 1)

 

C++




// Efficient C++ method to find sum of cubes of
// first n odd numbers.
#include <iostream>
using namespace std;
  
int cubeSum(int n)
{
    return n * n * (2 * n * n - 1);
}
  
int main()
{
    cout << cubeSum(4);
    return 0;
}


Java




// Java program to perform sum of
// cubes of first n odd natural numbers
  
public class GFG 
{
    public static int cubesum(int n)
    {
                  
        return (n) * (n) * (2 * n * n - 1);
    }
      
  
    // Driver function
    public static void main(String args[])
    {
        int a = 4;
        System.out.println(cubesum(a));
          
    }
}
  
// This code is contributed by Akansh Gupta.


Python3




# Python3 program to find sum of
# cubes of first n odd numbers.
  
# Function to find sum of cubes 
# of first n odd number 
def cubeSum(n):
    return (n * n * (2 * n * n - 1))
  
# Driven code 
print(cubeSum(4))
  
# This code is contributed by Shariq Raza


C#




// C# program to perform sum of
// cubes of first n odd natural numbers
using System;
  
public class GFG 
{
    public static int cubesum(int n)
    {
                  
        return (n) * (n) * (2 * n * n - 1);
    }
      
  
    // Driver function
    public static void Main()
    {
        int a = 4;
        Console.WriteLine(cubesum(a));
          
    }
}
  
// This code is published vt_m.


PHP




<?php
// Efficient PHP method to 
// find sum of cubes of
// first n odd numbers.
  
function cubeSum($n)
{
    return $n * $n * (2 * $n * $n - 1);
}
  
// Driver Code
echo cubeSum(4);
  
// This code is contributed by vt_m.
?>


Javascript




<script>
// javascript program to perform sum of
// cubes of first n odd natural numbers
  
function cubesum(n)
{
              
    return (n) * (n) * (2 * n * n - 1);
}
  
// Driver function
var a = 4;
document.write(cubesum(a));
  
// This code is contributed by Amit Katiyar 
</script>


Output: 
 

496

Complexity Analysis:

Time Complexity: O(1)

Space Complexity: O(1)

 



Last Updated : 16 Feb, 2023
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