查找给定系列的第n个项

给定该系列的前两个项分别为1和6，并且该系列的所有元素都比其前后的均值小2。任务是打印系列的^第n^个术语。
该系列的前几个术语是：

1, 6, 15, 28, 45, 66, 91, …

例子：

Input: N = 3 
Output: 15
Input: N = 1 
Output: 1 
 

方法：给定的级数表示三角数列中的奇数位置数。由于可以通过(n *(n + 1)/ 2)轻松找到^第n^个三角数，因此为了找到奇数，我们可以将n替换为(2 * n)– 1，因为(2 * n)– 1将总是产生奇数，即给定序列的^第n个数将是((2 * n)– 1)* n 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the nth term
// of the given series
int oddTriangularNumber(int N)
{
    return (N * ((2 * N) - 1));
}
 
// Driver code
int main()
{
    int N = 3;
    cout << oddTriangularNumber(N);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the nth term
// of the given series
static int oddTriangularNumber(int N)
{
    return (N * ((2 * N) - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
    System.out.println(oddTriangularNumber(N));
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python 3 implementation of the approach
 
# Function to return the nth term
# of the given series
def oddTriangularNumber(N):
    return (N * ((2 * N) - 1))
 
# Driver code
if __name__ == '__main__':
    N = 3
    print(oddTriangularNumber(N))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the nth term
    // of the given series
    static int oddTriangularNumber(int N)
    {
        return (N * ((2 * N) - 1));
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int N = 3;
        Console.WriteLine(oddTriangularNumber(N));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

// Javascript implementation of the approach
 
// Function to return the nth term
// of the given series
function oddTriangularNumber(N)
{
    return (N * ((2 * N) - 1));
}
 
// Driver code
let N = 3;
document.write(oddTriangularNumber(N));
 
// This code is contributed by subham348.

输出：

15