Find the Nth term of the series 9, 45, 243,1377

Given an integer N, the task is to print the N^th term of the series 9, 45, 243, 1377, 8019, …
Examples:

Input: N = 3 
Output: 243
Input: N = 5 
Output: 8019 

Approach: Let the N^th term be A_n, we can get the N^th term easily by observing the series: 

9, 45, 243, 1377, 8019, … 
(1¹ + 2¹) * 3¹, (1² + 2²) * 3², (1³ + 2³) * 3³, (1⁴ + 2⁴) * 3⁴, …, (1ⁿ + 2ⁿ) * 3ⁿ
So, A_n = (1ⁿ + 2ⁿ) * 3ⁿ

Algorithm:

Step 1: Start
step 2: create a function of static type with int return type with an integer parameter as input.
Step 3: Now we will use the pow function to raise the numbers 1, 2, and 3 to the power of n.
Step 4: Now it will return the value of the nth term.
Step 5: End

Below is the implementation of the above approach:

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the nth term of the given series
long nthterm(int n)
{

    // nth term
    int An = (pow(1, n) + pow(2, n)) * pow(3, n);

    return An;
}

// Driver code
int main()
{
    int n = 3;

    cout << nthterm(n);

    return 0;
}

Python

# Python3 implementation of the approach

# Function to return the nth term of the given series
def nthterm(n):

    # nth term
    An = (1**n + 2**n) * (3**n)

    return An;


# Driver code
n = 3
print(nthterm(n))

Java

// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;

public class GFG {

    // Function to return the nth term of the given series
    static int nthTerm(int n)
    {
        int An
            = ((int)Math.pow(1, n) + (int)Math.pow(2, n))
              * (int)Math.pow(3, n);

        return An;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 3;
        System.out.println(nthTerm(n));
    }
}

C#

// C# implementation of the approach
using System;
public class GFG {

    // Function to return the nth term of the given series
    static int nthTerm(int n)
    {

        int An
            = ((int)Math.Pow(1, n) + (int)Math.Pow(2, n))
              * (int)Math.Pow(3, n);

        return An;
    }

    // Driver code
    public static void Main()
    {
        int n = 3;
        Console.WriteLine(nthTerm(n));
    }
}

PHP

<?php
// PHP implementation of the approach

// Function to return the nth term of the given series
function nthterm($n)
{

    $An = (pow(1, $n) + pow(2, $n)) * pow(3, $n);

    // nth term of the given series
    return $An;

}

// Driver code
$n = 3;
echo nthterm($n);
?>

Javascript

<script>
// Javascript implementation of the approach

// Function to return the nth term of the given series
function nthterm(n)
{

    // nth term
    let An = (Math.pow(1, n) + Math.pow(2, n)) * Math.pow(3, n);

    return An;
}

// Driver code
let n = 3;

document.write(nthterm(n));

// This code is contributed by subhammahato348.
</script>

Output:

243

Time Complexity: O(log n)

Auxiliary Space: O(1)