Suma del cuadrado de los primeros n números pares

Dado un número n, hallar la suma de los cuadrados de los primeros n números naturales pares. 

Ejemplos: 

Input : 3
Output : 56 
22 + 42 + 62 = 56

Input : 8
Output : 816
22 + 42 + 62 + 82 + 102 + 122 + 142 + 162 

Una solución simple es atravesar n números pares y encontrar la suma de los cuadrados.  

// Simple C++ method to find sum
// of square of first n even numbers.
#include <iostream>
using namespace std;
 
int squareSum(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; i++)
        sum += (2 * i) * (2 * i);
    return sum;
}
 
// Driver Code
int main()
{
    cout << squareSum(8);
    return 0;
}

// Simple Java method to find sum of
// square of first n even numbers.
import java.io.*;
 
class GFG
{
    static int squareSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum += (2 * i) * (2 * i);
        return sum;
    }
     
    // Driver Code
    public static void main(String args[])
                  throws IOException
    {
        System.out.println(squareSum(8));
    }
}
 
// This code is contributed by Nikita Tiwari

# Simple Python3 code to
# find sum of square of
# first n even numbers
 
def squareSum( n ):
    sum = 0
    for i in range (0, n + 1):
        sum += (2 * i) * (2 * i)
         
    return sum
 
# driver code
ans = squareSum(8)
print (ans)
 
# This code is contributed by Saloni Gupta

// Simple C# method to find sum of
// square of first n even numbers.
using System;
 
class GFG
{
    static int squareSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum += (2 * i) * (2 * i);
         
        return sum;
    }
     
    // Driver code
    public static void Main()
    {
        Console.Write(squareSum(8));
    }
}
 
// This code is contributed by vt_m.

<?php
// Simple PHP method to find sum of
// square of first n even numbers.
 
function squareSum($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += (2 * $i) * (2 * $i);
    return $sum;
}
 
 
    echo squareSum(8);
 
 
// This code is contributed by vt_m.
?>

<script>
 
// Simple Javascript method to find sum
// of square of first n even numbers.
function squareSum(n)
{
    sum = 0;
    for(let i = 1; i <= n; i++)
        sum += (2 * i) * (2 * i);
         
    return sum;
}
 
// Driver Code
document.write(squareSum(8));
 
// This code is contributed by souravmahato348
 
</script>

Producción: 

816

Una solución eficiente es aplicar la siguiente fórmula.

sum = 2 * n * (n + 1) * (2 * n + 1)/3

How does it work? 
We know that sum of square of first 
n natural numbers is = n(n+1)/2

Sum of square of first n even natural numbers = 
        22 + 42 + .... + (2n)2
      = 4 * (12 + 22 + .... + n2)
      = 4 * n(n+1)(2n+1) / 6
      = 2 * n(n+1)(2n+1)/3 

// Efficient C++ method to find sum
// of square of first n even numbers.
#include <iostream>
using namespace std;
 
int squareSum(int n)
{
    return 2 * n * (n + 1) *
            (2 * n + 1) / 3;
}
 
// Driver code
int main()
{
    cout << squareSum(8);
    return 0;
}

// Efficient Java method to find sum
// of square of first n even numbers.
import java.io.*;
 
class GFG
{
    static int squareSum(int n)
    {
        return 2 * n * (n + 1) *
                (2 * n + 1) / 3;
    }
 
    // Driver Code
    public static void main(String args[])
                    throws IOException
    {
        System.out.println(squareSum(8));
    }
}
 
// This code is contributed by Nikita Tiwari

# Efficient Python3 code to
# find sum of square of
# first n even numbers
 
def squareSum( n ):
     
    return int(2 * n * (n + 1) * (2 * n + 1) / 3)
 
# driver code
ans = squareSum(8)
print (ans)
 
# This code is contributed by Saloni Gupta

// Efficient C# method to find sum
// of square of first n even numbers.
using System;
 
class GFG
{
     
    static int squareSum(int n)
    {
        return 2 * n * (n + 1) * (2 * n + 1) / 3;
    }
     
    // driver code
    public static void Main()
    {
        Console.Write(squareSum(8));
    }
}
 
// This code is contributed by vt_m.

<?php
// Efficient PHP method to find sum of
// square of first n even numbers.
 
 
function squareSum( $n)
{
    return 2 * $n * ($n + 1) *
             (2 * $n + 1) / 3;
}
 
echo squareSum(8);
 
// This code is contributed by vt_m.
?>

<script>
 
// JavaScript program to find sum of
// square of first n even numbers.
 
    function squareSum(n)
    {
        let sum = 0;
        for (let i = 1; i <= n; i++)
            sum += (2 * i) * (2 * i);
        return sum;
    }
 
// Driver code
 
            document.write(squareSum(8));
 
</script>

Salida : 

816

Complejidad temporal: O(1) desde que se realizan operaciones constantes