Sum of cubes of first n even numbers (original) (raw)

Last Updated : 16 Feb, 2023

Given a number n, find the 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 += (2i) * (2i) * (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

JavaScript

`

Output:

10368

Time Complexity: O(n)
Auxiliary Space: O(1)

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

Example

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 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

JavaScript

`

Output:

10368

Time Complexity: O(1)

Auxiliary Space: O(1)

Sum of cube of first n odd natural numbers