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Sum of cubes of first n even numbers
  • Difficulty Level : Easy
  • Last Updated : 24 Mar, 2021
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Given a number n, find sum of first n even natural numbers. 
Examples: 
 

Input : 2
Output : 72
2^3 + 4^3 = 72

Input : 8
Output :10368
2^3 + 4^3 + 6^3 + 8^3 + 10^3 + 12^3 + 14^3 + 16^3 = 10368

 

A simple solution is to traverse through n even numbers and find the sum of cubes. 
 

C++




// Simple C++ method to find sum of cubes of
// first n even numbers.
#include <iostream>
using namespace std;
 
int cubeSum(int n)
{
    int sum = 0;
    for (int i = 1; i <=  n; i++)
        sum += (2*i) * (2*i) * (2*i);
    return sum;
}
 
int main()
{
    cout << cubeSum(8);
    return 0;
}

Java




// Java program to perform
// sum of cubes of first
// n even natural numbers
 
public class GFG
{
    public static int cubesum(int n)
    {
        int sum = 0;
        for(int i = 1; i <= n; i++)
            sum += (2 * i) * (2 * i)
                   * (2 * i);
                 
        return sum;
    }
     
 
    // Driver function
    public static void main(String args[])
    {
        int a = 8;
        System.out.println(cubesum(a));
         
    }
}
 
// This code is contributed by Akansh Gupta

Python3




# Python3 program to find sum of
# cubes of first n even numbers
 
# Function to find sum of cubes
# of first n even numbers
def cubeSum(n):
 
    sum = 0
    for i in range(1, n + 1):
        sum += (2 * i) * (2 * i) * (2 * i)
    return sum
 
# Driven code
print(cubeSum(8))
 
# This code is contributed by Shariq Raza

C#




// C# program to perform
// sum of cubes of first
// n even natural numbers
using System;
 
public class GFG
{
    public static int cubesum(int n)
    {
        int sum = 0;
        for(int i = 1; i <= n; i++)
            sum += (2 * i) * (2 * i)
                * (2 * i);
                 
        return sum;
    }
     
 
    // Driver function
    public static void Main()
    {
        int a = 8;
        Console.WriteLine(cubesum(a));
         
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// Simple PHP method to
// find sum of cubes of
// first n even numbers.
 
function cubeSum($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += (2 * $i) *
                (2 * $i) *
                (2 * $i);
    return $sum;
}
 
// Driver Code
echo cubeSum(8);
     
// This code is contributed by vt_m.
?>

Javascript




<script>
// JavaScript program to find sum of cubes of
// first n even numbers.
 
    function cubeSum(n)
    {
        let sum = 0;
        for (let i = 1; i <=  n; i++)
        sum += (2*i) * (2*i) * (2*i);
        return sum;
    }
   
    document.write(cubeSum(8)); 
 
// This code is contributed by Surbhi Tyagi
 
</script>

Output:

10368

An efficient solution is to apply below formula.
 



sum = 2 *  n2(n+1)2 
 

How does it work? 

We know that sum of cubes of first 
n natural numbers is = n2(n+1)2 / 4

Sum of cubes of first n natural numbers = 
                2^3 + 4^3 + .... + (2n)^3
              = 8 * (1^3 + 2^3 + .... + n^3)
              = 8 *  n2(n+1)2 / 4
              = 2 *  n2(n+1)2 

 

C++




// Efficient C++ method to find sum of cubes of
// first n even numbers.
#include <iostream>
using namespace std;
 
int cubeSum(int n)
{
    return 2 * n * n * (n + 1) * (n + 1);
}
 
int main()
{
    cout << cubeSum(8);
    return 0;
}

Java




// Java program to perform
// sum of cubes of first
// n even natural numbers
 
public class GFG
{
    public static int cubesum(int n)
    {
                 
        return 2 * n * n * (n + 1) * (n + 1);
    }
     
 
    // Driver function
    public static void main(String args[])
    {
        int a = 8;
        System.out.println(cubesum(a));
         
    }
}
 
// This code is contributed by Akansh Gupta

Python3




# Python3 program to find sum of
# cubes of first n even numbers
 
# Function to find sum of cubes
# of first n even numbers
def cubeSum(n):
     
    return 2 * n * n * (n + 1) * (n + 1)
 
# Driven code
print(cubeSum(8))
 
# This code is contributed by Shariq Raza

C#




// C# program to perform
// sum of cubes of first
// n even natural numbers
using System;
 
class GFG
{
    public static int cubesum(int n)
    {
        return 2 * n * n *
               (n + 1) * (n + 1);
    }
     
    // Driver code
    public static void Main()
    {
        int a = 8;
        Console.WriteLine(cubesum(a));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// Efficient PHP code to
// find sum of cubes of
// first n even numbers.
 
function cubeSum($n)
{
    return 2 * $n * $n *
           ($n + 1) * ($n + 1);
}
 
// Driver code
echo cubeSum(8);
     
// This code is contributed by vt_m.
?>

Javascript




<script>
// javascript program to perform
// sum of cubes of first
// n even natural numbers
function cubesum(n)
{    
    return 2 * n * n * (n + 1) * (n + 1);
}
 
// Driver function
var a = 8;
document.write(cubesum(a));
 
// This code is contributed by Amit Katiyar
</script>

Output:

10368

Sum of cube of first n odd natural numbers
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.
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