找到系列 1, 3, 7, 15, 31 的第 N 项。 . .

给定一个正整数N，任务是找到序列的第 N 项：

1, 3, 7, 15, 31, …..

例子：

Input: N = 5
Output: 31

Input: N = 1
Output: 1

方法：

该序列是通过使用以下模式形成的。对于任何值 N-

T_N = 2^N – 1

插图：

Input: N = 5
Output: 31
Explanation:
T_N = 2^N – 1
     = 2⁵ – 1
     = 32 – 1 
     = 31

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to return
// Nth term of the series
int findTerm(int N)
{
    return pow(2, N) - 1;
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << findTerm(N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
public class GFG
{
 
  // Function to return
  // Nth term of the series
  static int findTerm(int N)
  {
    return (int)Math.pow(2, N) - 1;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 5;
    System.out.println(findTerm(N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python

# Python program to implement
# the above approach
 
# Function to return
# Nth term of the series
def findTerm(N):
     
    return pow(2, N) - 1
 
# Driver Code
N = 5
print(findTerm(N))
 
# This code is contributed by samim2000.

C#

// C# program to implement
// the above approach
using System;
 
class GFG
{
 
  // Function to return
  // Nth term of the series
  static int findTerm(int N)
  {
    return (int)Math.Pow(2, N) - 1;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 5;
    Console.Write(findTerm(N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript program to implement
// the above approach
 
// Function to return
// Nth term of the series
function findTerm(N)
{
    return Math.pow(2, N) - 1;
}
 
// Driver Code
let N = 5;
document.write(findTerm(N));
 
// This code is contributed by saurabh_jaiswal.

输出

31

时间复杂度： O(1)

辅助空间： O(1)