前n个偶数的平方和

给定一个数n，求第n个偶数(自然数)的平方和。

例子 ：

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

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

一个简单的解决方案是遍历n个偶数并找到平方和。

C++

// Simple C++ method to find sum
// of square of first n even numbers.
#include 
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;
}

Java

// 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

Python3

# 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

C#

// 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.

Javascript

 
// 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
 

输出：

816

一个有效的解决方案是应用以下公式。

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 

C++

// Efficient C++ method to find sum
// of square of first n even numbers.
#include 
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;
}

Java

// 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

Python3

# 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

C＃

// 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.

Java脚本

 
// 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));
 

输出：

816