查找系列的第N个项，其中每个项交替相差6和2

给定数字N ，任务是找到级数的第N个项，其中每个项交替相差6和2。
例子：

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 
 

天真的方法：这个想法是从1开始以6和2的增量迭代，直到达到第N个项。
下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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

 
// 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
 

输出：

14

时间复杂度： O(N)
高效的方法：我们可以使用以下公式找到第N个项：

1. 如果N为奇数：第N个项由(N / 2 +1)* 6 +(N / 2)* 2给出。
2. 如果N为偶数：第N个项由(N / 2)* 6 +(N / 2)* 2给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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

Java脚本

 
// 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
 

输出：

14

时间复杂度： O(1)