Programa para hallar el N-ésimo término de la serie 3, 7, 13, 21, 31…..

Dado un número N, la tarea es encontrar el N-ésimo término de esta serie:

3, 7, 13, 21, 31, …….

Ejemplos:

Input: N = 4
Output: 21
Explanation:
Nth term = (pow(N, 2) + N + 1)
         = (pow(4, 2) + 4 + 1)
         = 21

Input: N = 11
Output: 133

Enfoque:
$Sum = 0+3+7+13+21+31+........+a_{n-1} + a_n\\ Sum = 3+7+13+21+31+...+a_{n-2}+a_{n-1}+a_n$
restando estas dos ecuaciones obtenemos
$0=3+\left \{\frac{n-1}{2}\right \}[2*4 + (n-2)*2]-a_n\\ =3+\left \{\frac{n-1}{2}\right \}[8 + 2n-4]-a_n\\ =3+\left \{\frac{n-1}{2}\right \}[2n+4]-a_n\\ a_n=3+(n-1)(n+2)\\ a_n=n^2+n+1$
Por lo tanto, el N-ésimo término de la serie dada es:

A continuación se muestra la implementación del enfoque anterior:

C++

// CPP program to find the Nth term of given series.
#include <iostream>
#include <math.h>
using namespace std;
 
// Function to calculate sum
long long int getNthTerm(long long int N)
{
    // Return Nth term
    return (pow(N, 2) + N + 1);
}
 
// driver code
int main()
{
    // declaration of number of terms
    long long int N = 11;
 
    // Get the Nth term
    cout << getNthTerm(N);
 
    return 0;
}

Java

// Java code to find the Nth term of given series.
import java.util.*;
 
class solution
{
 
// Function to calculate sum
static long getNthTerm(long N)
{
     
   // Return Nth term
    return ((int)Math.pow(N, 2) + N + 1);
}
 
//Driver program
public static void main(String arr[])
{
     
   // declaration of number of terms
    long N = 11;
 
    // Get the Nth term
    System.out.println(getNthTerm(N));
 
}
}
//THis code is contributed by
//Surendra_Gangwar

Python3

# Python3 Code to find the
# Nth term of given series.
 
# Function to calculate sum
def getNthTerm(N):
     
    # Return Nth term
    return (pow(N, 2) + N + 1)
 
# driver code
if __name__=='__main__':
     
# declaration of number of terms
    N = 11
     
# Get the Nth term
    print(getNthTerm(N))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# code to find the Nth
// term of given series.
using System;
 
class GFG
{
 
// Function to calculate sum
static long getNthTerm(long N)
{
     
// Return Nth term
    return ((int)Math.Pow(N, 2) + N + 1);
}
 
// Driver Code
static public void Main ()
{
     
    // declaration of number
    // of terms
    long N = 11;
 
    // Get the Nth term
    Console.Write(getNthTerm(N));
}
}
 
// This code is contributed by Raj

PHP

<?php
// PHP program to find the
// Nth term of given series
 
// Function to calculate sum
function getNthTerm($N)
{
    // Return Nth term
    return (pow($N, 2) + $N + 1);
}
 
// Driver code
 
// declaration of number of terms
$N = 11;
 
// Get the Nth term
echo getNthTerm($N);
 
// This code is contributed by Raj
?>

Javascript

<script>
// JavaScript program to find the Nth term of given series.
 
// Function to calculate sum
function getNthTerm(N)
{
    // Return Nth term
    return (Math.pow(N, 2) + N + 1);
}
   
// driver code
 
   // declaration of number of terms
   let N = 11;
   
   // Get the Nth term
   document.write(getNthTerm(N));
   
// This code is contributed by Surbhi Tyagi
 
</script>

Producción:

Complejidad de tiempo: O(1)

Complejidad espacial: O(1) ya que usa variables constantes

Publicación traducida automáticamente

Artículo escrito por ash264 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

C++

Java

Python3

C#

PHP

Javascript

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