第 n 个五边形数

给定一个整数 n，找出第 n 个五边形数。前三个五边形数字是 1、5 和 12（请参见下图）。
第 n 个五边形数 P _n是由规则五边形的轮廓组成的点图案中不同点的数量，当五边形重叠时，边最多为 n 个点，因此它们共享一个顶点 [来源 Wiki]
例子 ：

Input: n = 1
Output: 1

Input: n = 2
Output: 5

Input: n = 3
Output: 12

通常，多边形数（三角形数、正方形数等）是表示为以正多边形形状排列的点或卵石的数。前几个五边形数字是：1、5、12等。
如果 s 是多边形的边数，则第 n 个 s 边数 P (s, n) 的公式为

nth s-gonal number P(s, n) = (s - 2)n(n-1)/2 + n

If we put s = 5, we get

n'th Pentagonal number Pn = 3*n*(n-1)/2 + n

例子：

五角数

下面是上述思想在不同编程语言中的实现。

C++

// C++ program for above approach
#include
using namespace std;
 
// Finding the nth pentagonal number
int pentagonalNum(int n)
{
    return (3 * n * n - n) / 2;
}
 
// Driver code
int main()
{
    int n = 10;
     
    cout << "10th Pentagonal Number is = "
         << pentagonalNum(n);
 
    return 0;
}
 
// This code is contributed by Code_Mech

C

// C program for above approach
#include 
#include 
 
// Finding the nth Pentagonal Number
int pentagonalNum(int n)
{
    return (3*n*n - n)/2;
}
 
// Driver program to test above function
int main()
{
    int n = 10;
    printf("10th Pentagonal Number is = %d \n \n",
                             pentagonalNum(n));
 
    return 0;
}

Java

// Java program for above approach
class Pentagonal
{
    int pentagonalNum(int n)
    {
        return (3*n*n - n)/2;
    }
}
 
public class GeeksCode
{
    public static void main(String[] args)
    {
        Pentagonal obj = new Pentagonal();
        int n = 10;   
        System.out.printf("10th petagonal number is = "
                          + obj.pentagonalNum(n));
    }
}

Python

# Python program for finding pentagonal numbers
def pentagonalNum( n ):
    return (3*n*n - n)/2
#Script Begins
 
n = 10
print "10th Pentagonal Number is = ", pentagonalNum(n)
  
#Scripts Ends

C#

// C# program for above approach
using System;
 
class GFG {
     
    static int pentagonalNum(int n)
    {
        return (3 * n * n - n) / 2;
    }
 
    public static void Main()
    {
        int n = 10;
         
        Console.WriteLine("10th petagonal"
        + " number is = " + pentagonalNum(n));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript program for above approach
 
    function pentagonalNum(n)
    {
        return (3 * n * n - n) / 2;
    }
 
// Driver code to test above methods
 
        let n = 10;
           
        document.write("10th petagonal"
        + " number is = " + pentagonalNum(n));
          
         // This code is contributed by avijitmondal1998.

输出 ：

10th Pentagonal Number is = 145

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