程序以找到系列0、5、18、39、67、105、150、203，…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的第N个术语：

0, 5, 18, 39, 67, 105, 150, 203, …(N Terms)

例子：

Input: N = 4
Output: 82
For N = 4
4th Term = ( 4 * 4 * 4 - 7 * 4 + 3) 
         = 39

Input: N = 10
Output: 332

方法：本系列的广义第N个术语：

$nth\: Term\: of\: the\: series\: = 4*n^2-7n+3$

以下是所需的实现：

C++

// CPP program to find the N-th term of the series:
// 0, 5, 18, 39, 67, 105, 150, 203, ...
#include 
#include 
using namespace std;
 
// calculate Nth term of series
int nthTerm(int n)
{
    return 4 * pow(n, 2) - 7 * n + 3;
}
 
// Driver code
int main()
{
    int N = 4;
 
    cout << nthTerm(N);
 
    return 0;
}

Java

// Java program to find the N-th term of the series:
// 0, 5, 18, 39, 67, 105, 150, 203,
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
// calculate Nth term of series
static int nthTerm(int n)
{
    return 4 * (int)Math.pow(n, 2) - 7 * n + 3;
}
   
// Driver code
public static void  main(String args[])
{
    int N = 4;
   
    System.out.print(nthTerm(N));
}
}

Python 3

# Python 3 program to find the
# N-th term of the series:
# 0, 5, 18, 39, 67, 105, 150, 203, ...
 
# calculate Nth term of series
def nthTerm(n):
 
    return 4 * pow(n, 2) - 7 * n + 3
 
# Driver code
N = 4
 
print(nthTerm(N))
 
# This code is contributed by ash264

C#

// C# program to find the
// N-th term of the series:
// 0, 5, 18, 39, 67, 105, 150, 203,
using System;
 
class GFG
{
// calculate Nth term of series
static int nthTerm(int n)
{
    return 4 * (int)Math.Pow(n, 2) -
                        7 * n + 3;
}
 
// Driver code
static public void Main ()
{
    int N = 4;
 
    Console.Write(nthTerm(N));
}
}
 
// This code is contributed by Raj

Javascript

 
// JavaScript program to find the N-th term of the series:
// 0, 5, 18, 39, 67, 105, 150, 203, ...
 
// calculate Nth term of series
function nthTerm( n)
{
    return 4 * Math.pow(n, 2) - 7 * n + 3;
}
// driver code
  // declaration of number of terms
   let N = 4;
   document.write( nthTerm(N) );
 
// This code contributed by gauravrajput1
 

输出：

39

时间复杂度： O(1)