Given a positive integer N. The task is to find Nth term of the series 3, 7, 19, 55, 163, …..
Examples:
Input: N = 5
Output: 163Input: N = 1
Output: 3
Approach: The sequence is formed by using the following pattern. For any value N
TN = 2 * 3N – 1 + 1
Illustration:
Input: N = 5
Output: 163
Explanation:
TN = 2 * 3N – 1 + 1
= 2 * 35 – 1 + 1
= 2 * 81 + 1
= 162 + 1
= 163
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to return Nth term // of the series int calcNum( int N) { return 2 * pow (3, N - 1) + 1; } // Driver Code int main() { int N = 5; cout << calcNum(N); return 0; } |
Java
// Java code to implement the above approach import java.lang.*; public class gfg { /* Function to return the Nth term of the series */ static int calcNum( int N) { return ( int )( 2 *(Math.pow( 3 ,N- 1 ))) + 1 ; } // Driver Code public static void main(String[] args) { int N = 5 ; System.out.println(calcNum(N)); } } // This code is contributed by Abhishek Thakur |
Python3
# Python3 program to implement # the above approach # Function to return Nth term # of the series def calcNum(N): return 2 * ( 3 * * (N - 1 )) + 1 # Driver Code N = 5 print (calcNum(N)) # This code is contributed by gfgking |
C#
// C# code to implement the above approach using System; public class gfg { /* Function to return the Nth term of the series */ static int calcNum( int N) { return ( int )(2 * (Math.Pow(3, N - 1))) + 1; } // Driver Code public static void Main( string [] args) { int N = 5; Console.WriteLine(calcNum(N)); } } // This code is contributed by Abhishek Thakur |
Javascript
<script> // JavaScript code for the above approach // Function to return Nth term // of the series function calcNum(N) { return 2 * Math.pow(3, N - 1) + 1; } // Driver Code let N = 5; document.write(calcNum(N)); // This code is contributed by Potta Lokesh </script> |
163
Time Complexity: O(logN) because using inbuilt pow function
Auxiliary Space: O(1)
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