Sum of cubes of first n odd natural numbers (original) (raw)

Last Updated : 16 Feb, 2023

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 += (2i + 1)(2i + 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)