Skip to content
Related Articles

Related Articles

Improve Article

Sum of cubes of first n odd natural numbers

  • Difficulty Level : Easy
  • Last Updated : 24 Mar, 2021

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

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++




// 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

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :