Find Nth term of the series 1, 8, 54, 384…

Given a number N. The task is to write a program to find the Nth term in the below series:

1, 8, 54, 384...

Examples:

Input : 3
Output : 54
For N = 3
Nth term = ( 3*3) * 3!
         = 54

Input : 2 
Output : 8

On observing carefully, the Nth term in the above series can be generalized as:



Nth term = ( N*N ) * ( N! )

Below is the implementation of the above approach:

C++

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// CPP program to find N-th term of the series:
// 1, 8, 54, 384...
#include <iostream>
using namespace std;
  
// calculate factorial of N
int fact(int N)
{
    int i, product = 1;
    for (i = 1; i <= N; i++)
        product = product * i;
    return product;
}
  
// calculate Nth term of series
int nthTerm(int N)
{
    return (N * N) * fact(N);
}
  
// Driver Function
int main()
{
    int N = 4;
  
    cout << nthTerm(N);
  
    return 0;
}

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Java

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// Java program to find N-th term of the series:
// 1, 8, 54, 384...
  
import java.io.*;
  
// Main class for main method
class GFG {
    public static int fact(int N)
    {
        int i, product = 1;
        // Calculate factorial of N
        for (i = 1; i <= N; i++)
            product = product * i;
        return product;
    }
    public static int nthTerm(int N)
    {
        // By using above formula
        return (N * N) * fact(N);
    }
  
    public static void main(String[] args)
    {
        int N = 4; // 4th term is 384
  
        System.out.println(nthTerm(N));
    }
}

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Python 3

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# Python 3 program to find 
# N-th term of the series:
# 1, 8, 54, 384...
  
# calculate factorial of N
def fact(N):
      
    product = 1
    for i in range(1, N + 1):
        product = product * i
    return product
  
# calculate Nth term of series
def nthTerm(N):
    return (N * N) * fact(N)
  
# Driver Code
if __name__ =="__main__":
    N = 4
    print(nthTerm(N))
  
# This code is contributed
# by ChitraNayal

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C#

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// C# program to find N-th 
// term of the series:
// 1, 8, 54, 384...
using System;
  
class GFG 
{
public static int fact(int N)
{
    int i, product = 1;
      
    // Calculate factorial of N
    for (i = 1; i <= N; i++)
        product = product * i;
    return product;
}
  
public static int nthTerm(int N)
{
    // By using above formula
    return (N * N) * fact(N);
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 4; // 4th term is 384
  
    Console.WriteLine(nthTerm(N));
}
}
  
// This code is contributed 
// by Kirti_Mangal

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PHP

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<?php 
// PHP program to find N-th 
/// term of the series:
// 1, 8, 54, 384...
  
// calculate factorial of N
function fact($N)
{
    $product = 1;
    for ($i = 1; $i <= $N; $i++)
        $product = $product * $i;
    return $product;
}
  
// calculate Nth term of series
function nthTerm($N)
{
    return ($N * $N) * fact($N);
}
  
// Driver Code
$N = 4;
  
echo nthTerm($N);
  
// This code is contributed 
// by ChitraNayal
?>

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Output:

384

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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Improved By : Kirti_Mangal, chitranayal