# Sum of cubes of first n odd natural numbers

• Difficulty Level : Easy
• Last Updated : 21 Jun, 2022

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 ``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

 ``

## Javascript

 ``

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 ``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

 ``

## Javascript

 ``

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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