程序找到系列2、12、28、50、77、112、152、198，…的Nth项。

给定数字N。任务是编写一个程序来查找以下系列的第N个术语：

2, 12, 28, 50, 77, 112, 152, 198…..(N Terms)

例子：

Input: N = 4
Output: 50
For N = 4
4th Term = ( 3 * 4 * 4 + 4 - 2) 
         = 50

Input: N = 10
Output: 307

方法：本系列的广义第N个术语：

$nth\: Term\: of\: the\: series\: = 3*n^2+n-2$

以下是所需的实现：

C++

// C++ program to find the N-th term of the series:
// 2, 12, 28, 50, 77, 112, 152, 198, .....
#include 
#include 
using namespace std;
 
// calculate Nth term of series
int nthTerm(int n)
{
    return 3 * pow(n, 2) + n - 2;
}
 
// Driver code
int main()
{
    int N = 4;
 
    cout << nthTerm(N) << endl;
 
    return 0;
}

Java

// Java program to find the N-th term of the series:
// 2, 12, 28, 50, 77, 112, 152, 198, .....
import java.util.*;
 
class solution
{
 
// calculate Nth term of series
static int nthTerm(int n)
{
return 3 *(int)Math.pow(n, 2) + n - 2;
}
 
//Driver program
public static void main(String arr[])
{
     
    int N = 4;
    System.out.println(nthTerm(N));
}
 
}
//This code is contributed by Surendra_Gangwar

Python3

# Python3 program to find the
# N-th term of the series:
# 2, 12, 28, 50, 77, 112, 152, 198, .....
 
# calculate Nth term of series
def nthTerm(n):
 
    return 3 * pow(n, 2) + n - 2
 
# Driver code
N = 4
print(nthTerm(N))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to find the
// N-th term of the series:
// 2, 12, 28, 50, 77, 112, 152, 198, .....
using System;
 
class GFG
{
 
// calculate Nth term of series
static int nthTerm(int n)
{
    return 3 * (int)Math.Pow(n, 2) +
                             n - 2;
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    Console.Write(nthTerm(N));
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

// JavaScript program to find the N-th term of the series:
// 2, 12, 28, 50, 77, 112, 152, 198, .....
 
// calculate Nth term of series
function nthTerm( n)
{
    return 3 * Math.pow(n, 2) + n - 2;
}
 
// Driver code
let N = 4;
 
   document.write( nthTerm(N) );
 
// This code contributed by gauravrajput1
 

输出：

50

时间复杂度： O(1)