Find Nth term of the series where each term differs by 6 and 2 alternately

Given a number N, the task is to find the Nth term of the series where each term differs by 6 and 2 alternately.
Examples:

Input: N = 6 
Output: 24 
Explanation:
The Nth term is 0 + 6 + 2 + 6 + 2 + 6 + 2 = 24
Input: N = 3 
Output: 14 
Explanation:
The Nth term is 0 + 6 + 2 + 6 = 14 

Naive Approach: The idea is to iterate from 1 with an increment of 6 and 2 alternatively, till we reach the Nth term.
Below is the implementation of the above approach: 

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find Nth term
void findNthTerm(int N)
{
    int ans = 0;

    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {

        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }

        // Else add 2
        else {
            ans = ans + 2;
        }
    }

    // Print ans
    cout << ans << endl;
}

// Driver Code
int main()
{
    int N = 3;
    findNthTerm(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find Nth term
static void findNthTerm(int N)
{
    int ans = 0;

    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {

        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }

        // Else add 2
        else {
            ans = ans + 2;
        }
    }

    // Print ans
    System.out.print(ans +"\n");
}

// Driver Code
public static void main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach

# Function to find Nth term
def findNthTerm(N):
    ans = 0

    # Iterate from 1 till Nth term
    for i in range(N):

        # Check if i is even and
        # then add 6
        if (i % 2 == 0) :
            ans = ans + 6


        # Else add 2
        else :
            ans = ans + 2

    # Print ans
    print(ans)


# Driver Code
if __name__=='__main__':

    N = 3
    findNthTerm(N)


# This code is contributed by AbhiThakur

C#

// C# program for the above approach
using System;

class GFG{

// Function to find Nth term
static void findNthTerm(int N)
{
    int ans = 0;

    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {

        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }

        // Else add 2
        else {
            ans = ans + 2;
        }
    }

    // Print ans
    Console.Write(ans +"\n");
}

// Driver Code
public static void Main()
{
    int N = 3;
    findNthTerm(N);
}
}

// This code is contributed by AbhiThakur

Javascript

<script>

// JavaScript program for the above approach

    // Function to find Nth term
    function findNthTerm(N)
    {
        let ans = 0;

        // Iterate from 1 till Nth term
        for (let i = 0; i < N; i++) {

            // Check if i is even and
            // then add 6
            if (i % 2 == 0) {
                ans = ans + 6;
            }

            // Else add 2
            else {
                ans = ans + 2;
            }
        }

        // Print ans
        document.write(ans + "<br>");
    }

    // Driver Code
       let N = 3;
       findNthTerm(N);


// This code is contributed by Mayank Tyagi

</script>

Output:

14

Time Complexity: O(N)

Auxiliary Space: O(1) as using constant variables

Efficient Approach: We can find the Nth term by using the below formula: 

1. If N is Odd: The Nth term is given by (N/2 + 1)*6 + (N/2)*2.
2. If N is Even: The Nth term is given by (N/2)*6 + (N/2)*2.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find Nth term
void findNthTerm(int N)
{
    int ans;

    // Check if N is even
    if (N % 2 == 0) {

        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }

    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }

    // Print ans
    cout << ans << endl;
}

// Driver Code
int main()
{
    int N = 3;
    findNthTerm(N);
    return 0;
}

Java

// Java program for the above approach


class GFG{

// Function to find Nth term
static void findNthTerm(int N)
{
    int ans;

    // Check if N is even
    if (N % 2 == 0) {

        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }

    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }

    // Print ans
    System.out.print(ans +"\n");
}

// Driver Code
public static void main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}

// This code contributed by PrinciRaj1992

Python3

# Python3 program for the above approach

# Function to find Nth term
def findNthTerm(N):
    ans = 0;

    # Check if N is even
    if (N % 2 == 0):

        # Formula for n is even
        ans = (N // 2) * 6 + (N // 2) * 2;

    # Check if N is odd
    else:

        # Formula for N is odd
        ans = (N // 2 + 1) * 6 + (N // 2) * 2;

    # Print ans
    print(ans);

# Driver Code
if __name__ == '__main__':

    N = 3;
    findNthTerm(N);

# This code is contributed by Rajput-Ji

C#

// C# program for the above approach
using System;

class GFG{

// Function to find Nth term
static void findNthTerm(int N)
{
    int ans;

    // Check if N is even
    if (N % 2 == 0) {

        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }

    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }

    // Print ans
    Console.Write(ans +"\n");
}

// Driver Code
public static void Main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program for the above approach

// Function to find Nth term
function findNthTerm( N)
{
    let ans;

    // Check if N is even
    if (N % 2 == 0) {

        // Formula for n is even
        ans = parseInt(N / 2) * 6
              + parseInt(N / 2) * 2;
    }

    // Check if N is odd
    else {
        // Formula for N is odd
        ans = parseInt(N / 2 + 1) * 6
              + parseInt(N / 2) * 2;
    }

    // Print ans
     document.write(ans);
}

// Driver Function

    // get the value of N
    let N = 3;

    // Calculate and print the Nth term
    findNthTerm(N);

// This code is contributed by todaysgaurav

</script>

Output:

14

Time Complexity: O(1)

Auxiliary Space: O(1) since using constant variables