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++ 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 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 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# 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
<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>
Output
163
Time Complexity: O(logN) because using inbuilt pow function
Auxiliary Space: O(1)