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Find the Nth term of the series 9, 45, 243,1377
• Last Updated : 08 Apr, 2021

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

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach``#include ``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

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## Javascript

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Output:
`243`

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