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

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

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

Approach: Let the Nth term be An, we can get the Nth 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, An = (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)



Last Updated : 26 Feb, 2023
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