Sum of cubes of first n even numbers

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.

// 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;
}
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 
// 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;
}
Output:10368

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