# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

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 ` `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

 ` `

Output:

`496`

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