程序以找到系列0、14、40、78、124等的第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, 14, 40, 78, 124 …(N Terms)

例子：

Input: N = 4
Output: 78
For N = 4
Sum(upto 4 terms) = ( 6 * n * n - 4 * n - 2) 
                  = ( 6 * 4 * 4 - 4 * 4 - 2) 
                  =  78
Input: N = 10
Output: 557

方法：本系列的广义第N个术语：
$nth\: Term\: of\: the\: series\: = 6*n^2-4n-2$
以下是所需的实现：

C++

// CPP program to find the N-th term of the series
// 0, 14, 40, 78, 124 ...
#include 
#include 
using namespace std;
 
// calculate sum upto Nth term of series
int nthTerm(int n)
{
    return 6 * pow(n, 2) - 4 * n - 2;
}
 
// 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, 14, 40, 78, 124 ...
import java.util.*;
class solution
{
 
// calculate sum up to Nth term of series
static int nthTerm(int n)
{
 
    //return the final sum
    return 6 * (int)Math.pow(n, 2) - 4 * n - 2;
}
 
// Driver code
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
# 0, 14, 40, 78, 124 ...
 
# calculate sum upto Nth
# term of series
def nthTerm(n):
    return int(6 * pow(n, 2) - 4 * n - 2)
 
# Driver code
N = 4
print(nthTerm(N))
 
# This code is contributed
# by Shrikant13

C#

// C# program to find the
// N-th term of the series
// 0, 14, 40, 78, 124 ...
using System;
 
class GFG
{
 
// calculate sum up to Nth
// term of series
static int nthTerm(int n)
{
 
    //return the final sum
    return 6 * (int)Math.Pow(n, 2) -
                        4 * n - 2;
}
 
// Driver code
public static void Main()
{
    int N = 4;
 
    Console.WriteLine(nthTerm(N));
}
}
 
// This code is contributed
// by inder_verma

---

Javascript

// JavaScript program to find the N-th term of the series
// 0, 14, 40, 78, 124 ...
 
// calculate sum upto Nth term of series
function nthTerm( n)
{
    return 6 * Math.pow(n, 2) - 4 * n - 2;
}
// Driver code
 
    let N = 4;
   document.write( nthTerm(N) );
 
// This code contributed by aashish1995
 

输出：

78

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