Given a number N. The task is to write a program to find the Nth term in the below series:
1, 4, 15, 72, 420…
Examples:
Input: 3
Output: 15
Explanation: For N = 3, we know that the factorial of 3 is 6 Nth term = 6*(3+2)/2Input: 6
Output: 2880
Explanation: For N = 6, we know that the factorial of 6 is 720 Nth term = 620*(6+2)/2 = 2880
The idea is to first find the factorial of the given number N, that is N!. Now the N-th term in the above series will be:
N-th term = N! * (N + 2)/2
Below is the implementation of the above approach:
C++
// CPP program to find N-th term of the series: // 1, 4, 15, 72, 420… #include <iostream> using namespace std; // Function to find factorial of N int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series int nthTerm(int N) { return (factorial(N) * (N + 2) / 2); } // Driver Function int main() { int N = 6; cout << nthTerm(N); return 0; } |
Java
// Java program to find N-th // term of the series: // 1, 4, 15, 72, 420 import java.util.*; import java.lang.*; import java.io.*; class GFG { // Function to find factorial of N static int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series static int nthTerm(int N) { return (factorial(N) * (N + 2) / 2); } // Driver Code public static void main(String args[]) { int N = 6; System.out.println(nthTerm(N)); } } // This code is contributed by Subhadeep |
Python3
# Python 3 program to find # N-th term of the series: # 1, 4, 15, 72, 420… # Function for finding # factorial of N def factorial(N) : fact = 1 for i in range(1, N + 1) : fact = fact * i # return factorial of N return fact# Function for calculating# Nth term of seriesdef nthTerm(N) : # return nth term return (factorial(N) * (N + 2) // 2)# Driver codeif __name__ == "__main__" : N = 6 # Function Calling print(nthTerm(N))# This code is contributed# by ANKITRAI1 |
C#
// C# program to find N-th // term of the series: // 1, 4, 15, 72, 420 using System; class GFG { // Function to find factorial of N static int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series static int nthTerm(int N) { return (factorial(N) * (N + 2) / 2); } // Driver Code public static void Main() { int N = 6; Console.Write(nthTerm(N)); } } // This code is contributed by ChitraNayal |
PHP
<?php// PHP program to find // N-th term of the series: // 1, 4, 15, 72, 420… // Function for finding // factorial of N function factorial($N){ $fact = 1; for($i = 1; $i <= $N; $i++) $fact = $fact * $i; // return factorial of N return $fact;}// Function for calculating// Nth term of seriesfunction nthTerm($N){ // return nth term return (factorial($N) * ($N + 2) / 2);}// Driver code$N = 6;// Function Callingecho nthTerm($N);// This code is contributed// by mits?> |
Javascript
<script>// JavaScript program to find N-th term of the series: // 1, 4, 15, 72, 420… // Function to find factorial of N function factorial( N) { let fact = 1; for (let i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series function nthTerm(N) { return (factorial(N) * (N + 2) / 2); } // Driver code let N = 6; document.write( nthTerm(N) );// This code contributed by aashish1995 </script> |
Output
2880
Time Complexity: O(n)
Auxiliary Space: O(1)
Another approach :(Using recursion)
C++
// CPP program to find N-th term of the series:// 1, 4, 15, 72, 420… // Using recursion#include <iostream>using namespace std;// Function to find factorial of N// with recursion int factorial(int N){ // base condition if( N == 0 || N == 1 ) return 1; // use recursion return N * factorial( N - 1 );}// calculate Nth term of seriesint nthTerm(int N){ return (factorial(N) * (N + 2) / 2);}// Driver Functionint main(){ int N = 6; cout << nthTerm(N); return 0;} |
Java
// Java program to find N-th // term of the series:// 1, 4, 15, 72, 420import java.util.*;import java.lang.*;import java.io.*;class GFG{ // Function to find factorial of Nstatic int factorial(int N){ // base condition if( N == 0 || N == 1 ) return 1; // use recursion return N * factorial( N - 1 );}// calculate Nth term of seriesstatic int nthTerm(int N){ return (factorial(N) * (N + 2) / 2);}// Driver Codepublic static void main(String args[]){ int N = 6; System.out.println(nthTerm(N));}} |
Python3
# Python3 program to find # N-th term of the series: # 1, 4, 15, 72, 420… # Using recursion # Function to find factorial # of N with recursiondef factorial(N): # base condition if N == 0 or N == 1: return 1 # use recursion return N * factorial(N - 1)def nthTerm(N): # calculate Nth term of series return (factorial(N) * (N + 2) // 2)# Driver codeN = 6print(nthTerm(N))# This code is contributed # by Shrikant13 |
C#
// C# program to find N-th // term of the series:// 1, 4, 15, 72, 420using System; class GFG{ // Function to find factorial of Nstatic int factorial(int N){ // base condition if( N == 0 || N == 1 ) return 1; // use recursion return N * factorial( N - 1 );}// calculate Nth term of seriesstatic int nthTerm(int N){ return (factorial(N) * (N + 2) / 2);}// Driver Codepublic static void Main(){ int N = 6; Console.Write(nthTerm(N));}}// This code is contributed by ChitraNayal |
PHP
<?php// PHP program to find// N-th term of the series:// 1, 4, 15, 72, 420… // Function to find factorial// of N with recursion function factorial($N){ // base condition if($N == 0 or $N == 1) return 1; // use recursion return $N * factorial($N - 1);}// calculate Nth term of seriesfunction nthTerm($N){ return (factorial($N) * ($N + 2) / 2);}// Driver Code$N = 6;echo nthTerm($N);// This code is contributed // by Shashank?> |
Javascript
<script>// Javascript program to find N-th// term of the series:// 1, 4, 15, 72, 420// Function to find factorial of Nfunction factorial(N){ // Base condition if (N == 0 || N == 1) return 1; // Use recursion return N * factorial(N - 1);}// Calculate Nth term of seriesfunction nthTerm(N){ return(factorial(N) * (N + 2) / 2);}// Driver Codelet N = 6;document.write(nthTerm(N));// This code is contributed by avanitrachhadiya2155</script> |
Output
2880
Time complexity: O(N)
Auxiliary Space: O(N), for recursion call stack.
Note: Above code wouldn’t work for large values of N. To find the values for large N, use the concept of Factorial for large numbers.
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