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C++ program to find Sum of cubes of first N Even numbers

Here, we are going to learn how to find the sum of the cubes of first N even numbers in C++ programming language?
Submitted by Vivek Kothari, on November 09, 2018

The problem is we have a number N and we have to find sum of first N Even natural numbers.

Example:

    Input:
    n = 3

    Output:
    288 (2^3 + 4^3+6^3)

A simple solution is given below...

Example 1:

#include <iostream> 
using namespace std; 

int calculate(int n) 
{ 
	int sum = 0; 
	for (int i = 1; i <=  n; i++) 
		sum = sum + (2*i) * (2*i) * (2*i); 
	return sum; 
} 

int main() 
{
	int num = 3;
	
	cout<<"Number is = "<<num<<endl; 
	cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num); 

	return 0; 
} 

Output

Number is = 3
Sum of cubes of first 3 even number is =288

The efficient approach is discussed below:

The sum of cubes of first n natural numbers is given by = (n*(n+1) / 2)^2
Sum of cubes of first n natural numbers can be written as...
= 2^3 + 4^3 + .... + (2n)^3
Now take out common term i.e 2^3 
    = 2^3 * (1^3 + 2^3 + .... + n^3)
    = 2^3*  (n*(n+1) / 2)^2
    = 8 * ((n^2)(n+1)^2)/4
    = 2 * n^2(n+1)^2

Now we can apply this formula directly to find the sum of cubes of first n even numbers.

Example 2:

#include <iostream> 
using namespace std; 

int calculate(int n) 
{
	int sum = 2 * n * n * (n + 1) * (n + 1);
	return sum; 
}   

int main() 
{
	int num = 3;
	cout<<"Number is = "<<num<<endl; 
	cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num); 
	return 0; 
} 

Output

Number is = 3
Sum of cubes of first 3 even number is =288




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