第二个十进制数

第二个十进制数系列可以表示为

7, 22, 45, 76, 115, 162, 217, 280,,…..

第N个学期
给定整数N。任务是找到给定级数的第N个项。
例子：

Input: N = 1 
Output: 7

Input: N = 4 
Output: 76   

方法：想法是找到第二个十进制数的通用术语。以下是第二个十进制数的通用项的计算：

1st Term = 1 * (4*1 + 3) = 7 
2nd term = 2 * (4*2 + 3) = 22 
3rd term = 3 * (4*3 + 3) = 45 
4th term = 4 * (4*4 + 3) = 76 
. 
. 
. 
Nth term = n * (4 * n + 3) 
Therefore, the Nth term of the series is given as $N * (4 * N + 3)$

下面是上述方法的实现：

C++

// C++ implementation to
// find N-th term in the series
 
#include 
#include 
using namespace std;
 
// Function to find N-th term
// in the series
void findNthTerm(int n)
{
    cout << n * (4 * n + 3) << endl;
}
 
// Driver Code
int main()
{
    int N = 4;
    findNthTerm(N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to find N-th term
// in the series
static void findNthTerm(int n)
{
    System.out.println(n * (4 * n + 3));
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
     
    findNthTerm(N);
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 implementation to
# find N-th term in the series
 
# Function to find N-th term
# in the series
def findNthTerm(n):
 
    print(n * (4 * n + 3))
 
# Driver Code
N = 4;
findNthTerm(N);
 
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find N-th term
// in the series
static void findNthTerm(int n)
{
    Console.WriteLine(n * (4 * n + 3));
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
     
    findNthTerm(N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript implementation to
// find N-th term in the series
 
// Function to find N-th term
// in the series
function findNthTerm(n)
{
    document.write(n * (4 * n + 3));
}
 
// Driver Code
let N = 4;
findNthTerm(N);
 
// This code is contributed by subhammahato348.

输出：

76

时间复杂度： O(1)

辅助空间： O(1)

参考： OEIS