Suma de cubos de los primeros n números naturales impares

Dado un número n, encuentre la suma de los primeros n números naturales impares.
 

Input  : 2
Output : 28
1^3 + 3^3 = 28

Input  : 4
Output : 496
1^3 + 3^3 + 5^3 + 7^3 = 496

Una solución simple es atravesar n números impares y encontrar la suma de los cubos. 
 

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

// 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
// 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 Code
echo 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>

Producción : 
 

28

Análisis de Complejidad:

Complejidad de tiempo: O(n), ya que estamos usando un solo recorrido en la función cubeSum().

Complejidad espacial:O(1)

Una solución eficiente es aplicar la siguiente fórmula.
 

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)

// 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 number
def cubeSum(n):
    return (n * n * (2 * n * n - 1))
 
# Driven code
print(cubeSum(4))
 
# This code is contributed by Shariq Raza

// 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
// Efficient PHP method to
// find sum of cubes of
// first n odd numbers.
 
function cubeSum($n)
{
    return $n * $n * (2 * $n * $n - 1);
}
 
// Driver Code
echo 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 function
var a = 4;
document.write(cubesum(a));
 
// This code is contributed by Amit Katiyar
</script>

Producción: 
 

496

Análisis de Complejidad:

Complejidad de tiempo: O(1)

Complejidad espacial: O(1)

Este artículo es una contribución de Dharmendra kumar . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.