Program to find the Nth term of the series 3, 7, 13, 21, 31.....

Given a number N, the task is to find the Nth term of this series:

3, 7, 13, 21, 31, .......

Examples: 

Input: N = 4
Output: 21
Explanation:
Nth term = (pow(N, 2) + N + 1)
         = (pow(4, 2) + 4 + 1)
         = 21
Input: N = 11
Output: 133

Approach:
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             
Subtracting these two equations we get 
 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            
Therefore, the Nth Term of the given series is: 

a_n=n^2+n+1

Below is the implementation of the above approach: 

// 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 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 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# 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

<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>

<?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
?>

133

Time Complexity: O(1)
Space Complexity: O(1) since using constant variables
 

Method 2: We can also solve the problem by the formula [ (n+1)²-n  ]. 

// 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 + 1, 2) - N);
}

// driver code
int main()
{
    // declaration of number of terms
    long long int N = 11;

    // Get the Nth term
    cout << getNthTerm(N);

    return 0;
}

// Nikunj Sonigara

public class Main {

    static long getNthTerm(long N) {
        return (long) (Math.pow(N + 1, 2) - N);
    }

    public static void main(String[] args) {
        long N = 11;
        System.out.println(getNthTerm(N));
    }
}

# Python program to find the Nth term of given series.
import math

# Function to calculate sum
def getNthTerm(N):
    # Return Nth term
    return int(math.pow(N + 1, 2) - N)

# driver code
if __name__ == '__main__':
    # declaration of number of terms
    N = 11

    # Get the Nth term
    print(getNthTerm(N))

using System;

class Program {
    // Function to calculate the Nth term of the series
    static long GetNthTerm(long N)
    {
        // Return Nth term
        return (long)(Math.Pow(N + 1, 2) - N);
    }

    static void Main()
    {
        // Declaration of the number of terms
        long N = 11;

        // Get the Nth term
        Console.WriteLine(GetNthTerm(N));
    }
}

// Nikunj Sonigara

function getNthTerm(N) {
    return Math.pow(N + 1, 2) - N;
}

const N = 11;
console.log(getNthTerm(N));

133

Time Complexity: O(logN)
Space Complexity: O(1) since using constant variables