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.
PHP
Javascript
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.
PHP
Javascript
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.
的PHP
Java脚本
输出:
816