求出前N个十二边形数的总和

给定数字N ，任务是找到前N个十二边形数的总和。

The first few dodecagonal numbers are 1, 12, 33, 64, 105, 156, 217 … 
 

例子：

Input: N = 3 
Output: 46 
Explanation: 
1, 12 and 33 are the first three Dodecagonal numbers
Input: N = 5 
Output: 215 
 

方法：

1. 最初，我们需要创建一个函数，该函数将帮助我们计算第N^个十二边形数。
2. 运行一个从1到N的循环，以找到第十二个对角线数。
3. 将所有以上计算得出的十二边形数相加。
4. 最后，显示前N个十二边形数字的总和。

下面是上述方法的实现：

C++

// C++ program to find the sum of
// the first N dodecagonal numbers
#include 
using namespace std;
 
// Function to find the N-th
// dodecagonal number
int Dodecagonal_num(int n)
{
 
    // Formula to calculate N-th
    // dodecagonal number
    return (5 * n * n - 4 * n);
}
 
// Function to find the sum of
// the first N dodecagonal numbers
int sum_Dodecagonal_num(int n)
{
 
    // Variable to get the sum
    int summ = 0;
 
    // Iterating through the
    // first N numbers
    for(int i = 1; i < n + 1; i++)
    {
         
        // Compute the sum
        summ += Dodecagonal_num(i);
    }
    return summ;
}
 
// Driver Code
int main()
{
    int n = 5;
 
    // Display first Nth
    // centered_decagonal number
    cout << (sum_Dodecagonal_num(n));
    return 0;
}
 
// This code is contributed by PrinciRaj1992

Java

// Java program to find the sum of
// the first N dodecagonal numbers
class GFG {
     
// Function to find the N-th
// dodecagonal number
static int Dodecagonal_num(int n)
{
 
    // Formula to calculate N-th
    // dodecagonal number
    return (5 * n * n - 4 * n);
}
 
// Function to find the sum of
// the first N dodecagonal numbers
static int sum_Dodecagonal_num(int n)
{
 
    // Variable to get the sum
    int summ = 0;
 
    // Iterating through the
    // first N numbers
    for(int i = 1; i < n + 1; i++)
    {
        
       // Compute the sum
       summ += Dodecagonal_num(i);
    }
    return summ;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 5;
 
    // Display first Nth
    // centered_decagonal number
    System.out.println(sum_Dodecagonal_num(n));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to find the
# sum of the first N
# Dodecagonal numbers
 
# Function to find the N-th
# Dodecagonal number
def Dodecagonal_num(n):
 
    # Formula to calculate 
    # N-th Dodecagonal
    # number 
    return (5 * n * n - 4 * n)
     
   
# Function to find the
# sum of the first N
# Dodecagonal numbers
def sum_Dodecagonal_num(n) :
     
    # Variable to get the sum
    summ = 0
     
    # Iterating through the
    # first N numbers
    for i in range(1, n + 1):
 
        # Compute the sum
        summ += Dodecagonal_num(i)
     
    return summ
   
# Driver Code
if __name__ == '__main__' :
           
    n = 5
     
    print(sum_Dodecagonal_num(n))

C#

// C# program to find the sum of
// the first N dodecagonal numbers
using System;
 
class GFG {
     
// Function to find the N-th
// dodecagonal number
static int Dodecagonal_num(int n)
{
 
    // Formula to calculate N-th
    // dodecagonal number
    return (5 * n * n - 4 * n);
}
 
// Function to find the sum of
// the first N dodecagonal numbers
static int sum_Dodecagonal_num(int n)
{
 
    // Variable to get the sum
    int summ = 0;
 
    // Iterating through the
    // first N numbers
    for(int i = 1; i < n + 1; i++)
    {
         
        // Compute the sum
        summ += Dodecagonal_num(i);
    }
    return summ;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 5;
 
    // Display first Nth
    // centered_decagonal number
    Console.WriteLine(sum_Dodecagonal_num(n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
    // Javascript program to find the sum of
    // the first N dodecagonal numbers
     
    // Function to find the N-th
    // dodecagonal number
    function Dodecagonal_num(n)
    {
 
        // Formula to calculate N-th
        // dodecagonal number
        return (5 * n * n - 4 * n);
    }
 
    // Function to find the sum of 
    // the first N dodecagonal numbers
    function sum_Dodecagonal_num(n)
    {
 
        // Variable to get the sum
        let summ = 0;
 
        // Iterating through the
        // first N numbers
        for(let i = 1; i < n + 1; i++)
        {
 
            // Compute the sum
            summ += Dodecagonal_num(i);
        }
        return summ;
    }
     
    let n = 5;
   
    // Display first Nth
    // centered_decagonal number
    document.write(sum_Dodecagonal_num(n));
 

输出：

215

时间复杂度： O(N)。