找到系列 5、10、20、40 的第 N 项…

给定一个正整数N ，任务是找到序列的第 N项

5, 10, 20, 40….till N terms

例子：

Input: N = 5
Output: 80

Input: N = 3
Output: 20

方法：

1st term = 5 * (2 ^ (1 – 1))  = 5

2nd term = 5 * (2 ^ (2 – 1)) = 10

3rd term = 5 * (2 ^ (3 – 1)) = 20

4th term = 5 * (2 ^ (4 – 1)) = 40

.

.

Nth term = 5 * (2 ^ (N – 1))

给定系列的第 N 项可以概括为-

T_N = (a * (r ^ (N – 1))

可以按照以下步骤推导出公式-

The series 5, 10, 20, 40….till N terms 

is in G.P. with 

first term a = 5

common ratio r = 2 because each term is double the one before it.

The Nth term of a G.P. is

T_N = (a * (r ^ (N – 1))

插图：

Input: N = 5
Output: 80
Explanation:
T_N = (a * (r ^ (N – 1))
     = (5 * (2 ^ (5 – 1))
     = (5 * 16)
     = 80

以下是上述方法的实现 -

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate nth term
int nTerm(int a, int r, int n)
{
    return a * pow(r, n - 1);
}
 
// Driver code
int main()
{
    // Value of N
    int N = 5;
 
    // First term of the series
    int a = 5;
 
    // Common ratio
    int r = 2;
 
    cout << nTerm(a, r, N);
    return 0;
}

C

// C program to implement
// the above approach
#include 
#include 
 
// Function to calculate nth term
int nTerm(int a, int r, int n)
{
    return a * pow(r, n - 1);
}
 
// Driver code
int main()
{
    // Value of N
    int N = 5;
 
    // First term
    int a = 5;
 
    // Common ratio
    int r = 2;
 
    printf("%d", nTerm(a, r, n));
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG {
    // Driver code
    public static void main(String[] args)
    {
        // Value of N
        int N = 5;
 
        // First term
        int a = 5;
 
        // Common ratio
        int r = 2;
        System.out.println(nTerm(a, r, N));
    }
 
    // Function to calculate nth term
    public static int nTerm(int a, int r, int n)
    {
        return a * ((int)Math.pow(r, n - 1));
    }
}

Python3

# python3 program to implement
# the above approach
 
# Function to calculate nth term
def nTerm(a, r, n):
 
    return a * pow(r, n - 1)
 
# Driver code
if __name__ == "__main__":
 
    # Value of N
    N = 5
 
    # First term of the series
    a = 5
 
    # Common ratio
    r = 2
 
    print(nTerm(a, r, N))
 
# This code is contributed by rakeshsahni

C#

using System;
 
public class GFG
{
   
    // Function to calculate nth term
    public static int nTerm(int a, int r, int n)
    {
        return a * ((int)Math.Pow(r, n - 1));
    }
    static public void Main()
    {
 
        // Code
        // Value of N
        int N = 5;
 
        // First term
        int a = 5;
 
        // Common ratio
        int r = 2;
        Console.Write(nTerm(a, r, N));
    }
}
 
// This code is contributed by Potta Lokesh

Javascript

      // JavaScript code for the above approach
 
      // Function to calculate nth term
      function nTerm(a, r, n) {
          return a * Math.pow(r, n - 1);
      }
 
      // Driver code
 
      // Value of N
      let N = 5;
 
      // First term of the series
      let a = 5;
 
      // Common ratio
      let r = 2;
 
      document.write(nTerm(a, r, N));
 
 
     // This code is contributed by Potta Lokesh
  

输出

80

时间复杂度： O(1)

辅助空间： O(1)