Encuentra el N-ésimo término de la serie 1, 2, 6, 21, 88, 445. . .

Dado un entero positivo N . La tarea es encontrar el término N de la serie:

1, 2, 6, 21, 88, 445, . . .

Ejemplos:

Entrada: N = 3
Salida: 6

Entrada: N = 6
Salida: 445

 

Acercarse: 

La secuencia dada sigue el siguiente patrón:

1, (1 * 1 + 1 = 2), (2 * 2 + 2 = 6), (6 * 3 + 3 = 21), (21 * 4 + 4 = 88), (88 * 5 + 5 = 445 ),… 

Los pasos a continuación se pueden usar para resolver el problema:

  • Para cada iterativo i , multiplicando su elemento anterior por i (Inicialmente el elemento será 1)
  • Y suma elementos multiplicados con i.
  • Finalmente, devuelva el N-ésimo término de la serie.

A continuación se muestra la implementación del enfoque anterior.

C++

// C++ program to find N-th term
// of the series-
// 1, 2, 6, 21, 88, 445...
#include <bits/stdc++.h>
using namespace std;
 
// Function to return Nth term
// of the series
int nthTerm(int N)
{
    // Initializing a variable
    int term = 1;
 
    // Loop to iterate from 1 to N-1
    for (int i = 1; i < N; i++)
    {
        // Multiplying and adding previous
        // element with i
        term = term * i + i;
    }
 
    // returning the Nth term
    return term;
}
 
// Driver Code
int main()
{
    // Get the value of N
    int N = 6;
    cout << nthTerm(N);
    return 0;
}

Java

// Java program to find N-th term
// of the series-
// 1, 2, 6, 21, 88, 445...
class GFG
{
   
    // Function to return Nth term
    // of the series
    static int nthTerm(int N)
    {
       
        // Initializing a variable
        int term = 1;
 
        // Loop to iterate from 1 to N-1
        for (int i = 1; i < N; i++)
        {
           
            // Multiplying and adding previous
            // element with i
            term = term * i + i;
        }
 
        // returning the Nth term
        return term;
    }
 
    // Driver Code
    public static void main(String args[])
    {
       
        // Get the value of N
        int N = 6;
        System.out.println(nthTerm(N));
    }
 
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# Python code for the above approach
 
# Function to return Nth term
# of the series
def nthTerm(N):
 
    # Initializing a variable
    term = 1;
 
    # Loop to iterate from 1 to N-1
    for i in range(1, N):
       
        # Multiplying and adding previous
        # element with i
        term = term * i + i;
 
    # returning the Nth term
    return term;
 
# Driver Code
 
# Get the value of N
N = 6;
print(nthTerm(N));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program to find N-th term
// of the series-
// 1, 2, 6, 21, 88, 445...
using System;
class GFG
{
 
  // Function to return Nth term
  // of the series
  static int nthTerm(int N)
  {
 
    // Initializing a variable
    int term = 1;
 
    // Loop to iterate from 1 to N-1
    for (int i = 1; i < N; i++)
    {
 
      // Multiplying and adding previous
      // element with i
      term = term * i + i;
    }
 
    // returning the Nth term
    return term;
  }
 
  // Driver Code
  public static void Main()
  {
 
    // Get the value of N
    int N = 6;
    Console.Write(nthTerm(N));
  }
 
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
        // JavaScript code for the above approach
 
        // Function to return Nth term
        // of the series
        function nthTerm(N)
        {
         
            // Initializing a variable
            let term = 1;
 
            // Loop to iterate from 1 to N-1
            for (let i = 1; i < N; i++)
            {
             
                // Multiplying and adding previous
                // element with i
                term = term * i + i;
            }
 
            // returning the Nth term
            return term;
        }
 
        // Driver Code
 
        // Get the value of N
        let N = 6;
        document.write(nthTerm(N));
 
  // This code is contributed by Potta Lokesh
    </script>
Producción

445

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por athakur42u y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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