📌  相关文章
📜  在系列12、35、81、173、357,…中找到第N个术语

📅  最后修改于: 2021-04-22 09:15:31             🧑  作者: Mango

给定数字N ,任务是找到序列12、35、81、173、357,…中第N个术语
例子:

Input: N = 2
Output: 35
2nd term = (12*2) + 11
         = 35

Input: N = 5
Output: 357
5th term = (12*(2^4))+11*((2^4)-1)
         = 357

方法:

  • 每个数字都是通过将前一个数字乘以2并加上11来获得的。
  • 由于起始编号为12。
1st term = 12
2nd term = (12 * 2) / 11 = 35
3rd term = (35 * 2) / 11 = 81
4th term = (81 * 2) / 11 = 173
And, so on....
  • 通常,第N个数字可通过以下公式获得:

下面是上述方法的实现:

C++
// C++ program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
 
#include 
using namespace std;
 
// Function to find Nth term
int nthTerm(int N)
{
    int nth = 0, first_term = 12;
 
    // Nth term
    nth = (first_term * (pow(2, N - 1)))
          + 11 * ((pow(2, N - 1)) - 1);
 
    return nth;
}
 
// Driver Method
int main()
{
    int N = 5;
    cout << nthTerm(N) << endl;
 
    return 0;
}


Java
// Java program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
class GFG
{
 
// Function to find Nth term
static int nthTerm(int N)
{
    int nth = 0, first_term = 12;
 
    // Nth term
    nth = (int) ((first_term * (Math.pow(2, N - 1)))
        + 11 * ((Math.pow(2, N - 1)) - 1));
 
    return nth;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 5;
    System.out.print(nthTerm(N) +"\n");
}
}
 
// This code is contributed by Rajput-Ji


Python3
# Python3 program to find the Nth term
# in series 12, 35, 81, 173, 357, ...
 
# Function to find Nth term
def nthTerm(N) :
 
    nth = 0; first_term = 12;
 
    # Nth term
    nth = (first_term * (pow(2, N - 1))) + \
            11 * ((pow(2, N - 1)) - 1);
 
    return nth;
 
# Driver Method
if __name__ == "__main__" :
 
    N = 5;
    print(nthTerm(N)) ;
 
# This code is contributed by AnkitRai01


C#
// C# program to find the Nth term
// in series 12, 35, 81, 173, 357, ...
using System;
 
class GFG
{
 
// Function to find Nth term
static int nthTerm(int N)
{
    int nth = 0, first_term = 12;
 
    // Nth term
    nth = (int) ((first_term * (Math.Pow(2, N - 1)))
        + 11 * ((Math.Pow(2, N - 1)) - 1));
 
    return nth;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 5;
    Console.Write(nthTerm(N) +"\n");
}
}
 
// This code is contributed by PrinciRaj1992


Javascript


输出:
357