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

Disclaimer: This content belongs to geeksforgeeks, source: http://geeksforgeeks.org```

rakesh