前N个Pronic数的总和

给定数字N ，任务是找到前N个Pronic数的总和。

The numbers that can be arranged to form a rectangle are called Rectangular Numbers (also known as Pronic numbers). The first few rectangular numbers are:
0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462 . . . .

例子：

Input: N = 4
Output: 20
Explanation:
0, 2, 6, 12 are the first 4 pronic numbers.

Input: N = 3
Output: 8

方法：

令，第N项由T _N表示。可以通过如下拆分每个术语来轻松解决此问题：

$S_N = 0 + 2 + 6 + 12 + 20 + 30......$
$S_N = (0 * 1) + (1 * 2) + (2 * 3) + (3 * 4) + (4 * 5) + ......$
$S_N = (0 * (0 + 1) + (1 * (1 + 1)) + (2 * (2 + 1)) + (3 * (3 + 1)) + (4 * (4 + 1)) + ......$
$S_N = (0^2 + 0) + (1^2 + 1) + (2^2 + 2) + (3^2 + 3) + (4^2 + 4) +......$
$S_N = (0 + 1 + 2 + 3 + 4 + ...T_N) + (0^2 + 1^2 + 2^2 + 3^2 + 4^2 + ... T_N)$

所以：

S_N = Sum of N Pronic Numbers
$S_N = [N*\frac{(N - 1)}{2}] + [N*\frac{(N - 1)*(2*N - 1)}{6}]$

下面是上述方法的实现：

C++

// C++ implementation to find 
// sum of first N terms
#include 
using namespace std;
  
// Function to calculate the sum
int calculateSum(int N)
{
  
    return N * (N - 1) / 2
           + N * (N - 1)
                 * (2 * N - 1) / 6;
}
  
// Driver code
int main()
{
    int N = 3;
  
    cout << calculateSum(N);
  
    return 0;
}

Java

// Java implementation implementation to find 
// sum of first N terms 
class GFG{ 
      
// Function to calculate the sum 
static int calculateSum(int N) 
{ 
      
    return N * (N - 1) / 2 + N * (N - 1) *
           (2 * N - 1) / 6; 
} 
      
// Driver code 
public static void main (String[] args) 
{ 
    int N = 3; 
      
    System.out.println(calculateSum(N)); 
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 implementation to find 
# sum of first N terms
  
# Function to calculate the sum
def calculateSum(N):
  
    return (N * (N - 1) // 2 + 
            N * (N - 1) * (2 * 
                 N - 1) // 6);
  
# Driver code
N = 3;
print(calculateSum(N));
  
# This code is contributed by Code_Mech

C#

// C# implementation implementation to find 
// sum of first N terms 
using System;
class GFG{ 
      
// Function to calculate the sum 
static int calculateSum(int N) 
{ 
      
    return N * (N - 1) / 2 + N * (N - 1) *
                        (2 * N - 1) / 6; 
} 
      
// Driver code 
public static void Main() 
{ 
    int N = 3; 
      
    Console.Write(calculateSum(N)); 
} 
} 
  
// This code is contributed by Code_Mech

输出：

8

时间复杂度： O(1)。