程序以找到系列0、5、14、27、44 ……..的第N个项。

给定数字N。任务是编写一个程序来查找以下系列中的第N个术语：

0, 5, 14, 27, 44 …(Nth Term)

例子：

Input: N = 4
Output: 27
For N = 4,
Nth term = ( 2 * N * N - N - 1 ) 
         = ( 2 * 4 * 4 - 4 - 1 ) 
         = 27

Input: N = 10
Output: 188

方法：本系列的广义第N个术语：

Nth Term: 2 * N * N - N - 1 

以下是所需的实现：

C++

// CPP program to find N-th term of the series:
// 0, 5, 14, 27, 44 ...
#include 
#include 
using namespace std;
 
// Calculate Nth term of series
int nthTerm(int n)
{
    return 2 * pow(n, 2) - n - 1;
}
 
// Driver code
int main()
{
    int N = 4;
 
    cout << nthTerm(N);
 
    return 0;
}

Java

// Java program to find N-th term of the series:
// 0, 5, 14, 27, 44 ...
import java.util.*;
 
class solution
{
 
// Calculate Nth term of series
static int nthTerm(int n)
{
    return 2 *(int)Math.pow(n, 2) - n - 1;
}
 
// Driver code
public static void main(String arr[])
{
    int N = 4;
 
    System.out.println(nthTerm(N));
}
}
//This code is contributed by Surendra_Gangwar

Python 3

# Python 3 program to find
# N-th term of the series:
# 0, 5, 14, 27, 44 ...
 
# Calculate Nth term of series
def nthTerm(n):
 
    return 2 * pow(n, 2) - n - 1
 
# Driver code
if __name__ == "__main__":
    N = 4
 
    print(nthTerm(N))
 
# This code is contributed
# by ChitraNayal

C#

// C# program to find
// N-th term of the series:
// 0, 5, 14, 27, 44 ...
using System;
class GFG
{
 
// Calculate Nth term of series
static int nthTerm(int n)
{
    return 2 * (int)Math.Pow(n, 2) - n - 1;
}
 
// Driver code
static public void Main ()
{
    int N = 4;
     
    Console.Write(nthTerm(N));
}
}
 
// This code is contributed by Raj

Javascript

// JavaScript program to find N-th term of the series:
// 0, 5, 14, 27, 44 ...
 
// Calculate Nth term of series
function nthTerm( n)
{
    return 2 * Math.pow(n, 2) - n - 1;
}
 
// Driver code
 
    let N = 4;
   document.write( nthTerm(N) );
 
// This code contributed by aashish1995
 

输出：

27

时间复杂度： O(1)
注意：以上系列(Sn)的n个项之和为：
$$ S_n = 2 \sum_{i=1}^n n^2 - \sum_{i=1}^n n -1\\ S_n=\frac{2n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}-n\\ S_n=\frac{n(n+1)(4n-1)-24}{6}\\ $$