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++

// 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)

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