Sum of cubes of first n odd natural numbers - GeeksforGeeks

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++
Java
Python3
C#
PHP
Javascript
// 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 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 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# 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
<?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 Codeecho cubeSum(2); // This code is contributed by vt_m.?>
<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++
Java
Python3
C#
PHP
Javascript
// 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 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 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# 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.
<?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 Codeecho cubeSum(4); // This code is contributed by vt_m.?>
<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 functionvar 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. 
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