# 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 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 codeprint(cubeSum(2)) # This code is contributed by Shariq Raza

## C#

 // C# program to perform sum of// cubes of first n odd natural numbersusing 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



## Javascript



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 numberdef cubeSum(n):    return (n * n * (2 * n * n - 1)) # Driven codeprint(cubeSum(4)) # This code is contributed by Shariq Raza

## C#

 // C# program to perform sum of// cubes of first n odd natural numbersusing 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.



## Javascript



Output:

496

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