# Sum of cubes of first n even numbers

• Difficulty Level : Easy
• Last Updated : 24 Mar, 2021

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 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 numbersdef cubeSum(n):     sum = 0    for i in range(1, n + 1):        sum += (2 * i) * (2 * i) * (2 * i)    return sum # Driven codeprint(cubeSum(8)) # This code is contributed by Shariq Raza

## C#

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



## Javascript



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 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 numbersdef cubeSum(n):         return 2 * n * n * (n + 1) * (n + 1) # Driven codeprint(cubeSum(8)) # This code is contributed by Shariq Raza

## C#

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



## Javascript



Output:

10368

Sum of cube of first n odd natural numbers
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