Sum of cubes of first n odd natural numbers

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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.
?>

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Output :

28

An efficient solution is to apply 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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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.

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PHP

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<?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.
?>

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Output:

496

This article is contributed by Dharmendra kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m