Find the sum of the first Nth Icosidigonal Number

Given a number N, the task is to find the sum of the first N Icosidigonal Numbers.

The first few Icosidigonal numbers are 1, 22, 63, 124, 205, 306 …

Examples:

Input: N = 3
Output: 86
Explanation:
1, 22 and 63 are the first three Icosidigonal numbers.

Input: N = 6
Output: 721



Approach:

  1. Initially, we need to create a function which will help us to calculate the Nth icosidigonal number.
  2. Now, run a loop starting from 1 to N, to find ith icosidigonal numbers.
  3. Add all the above calculated icosidigonal numbers.
  4. Finally, display the sum of the first N Icosidigonal numbers.

Below is the implementation of the above approach:

C++

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// C++ program to find the sum of the
// first N icosidigonal numbers
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the 
// N-th icosidigonal number 
int Icosidigonal_num(int n) 
{
      
    // Formula to calculate 
    // nth icosidigonal number
    return (20 * n * n - 18 * n) / 2;
}
  
// Function to find the sum of the
// first N icosidigonal number 
int sum_Icosidigonal_num(int n)
{
      
    // Variable to store the sum
    int summ = 0;
      
    // Iterating in the range 1 to N
    for(int i = 1; i < n + 1; i++)
    {
          
        // Finding the sum
        summ += Icosidigonal_num(i);
    }
    return summ;
}
  
// Driver code 
int main() 
    int n = 6;
      
    // Display first Nth 
    // icosidigonal numbers
    cout << sum_Icosidigonal_num(n);
  
// This code is contributed by coder001

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Java

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// Java program to find the sum of the 
// first N icosidigonal numbers 
class GFG{
      
// Function to find the 
// N-th icosidigonal number 
public static int Icosidigonal_num(int n) 
          
    // Formula to calculate 
    // nth icosidigonal number 
    return (20 * n * n - 18 * n) / 2
      
// Function to find the sum of the 
// first N icosidigonal number 
public static int sum_Icosidigonal_num(int n) 
          
    // Variable to store the sum 
    int summ = 0
          
    // Iterating in the range 1 to N 
    for(int i = 1; i < n + 1; i++) 
    
  
       // Finding the sum 
       summ += Icosidigonal_num(i); 
    
    return summ; 
  
// Driver code    
public static void main(String[] args)
{
    int n = 6
      
    // Display first Nth 
    // icosidigonal numbers 
    System.out.println(sum_Icosidigonal_num(n));
}
}
  
// This code is contributed by divyeshrabadiya07    

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Python3

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# Python3 program to find the 
# sum of the first N  
# Icosidigonal numbers
  
# Function to find the 
# N-th Icosidigonal 
# number 
def Icosidigonal_num(n):
   
    # Formula to calculate  
    # nth Icosidigonal 
    # number
    return (20 * n * n - 
            18 * n) // 2
      
    
# Function to find the 
# sum of the first N
# Icosidigonal number 
def sum_Icosidigonal_num(n) : 
      
    # Variable to store
    # the sum
    summ = 0
      
    # Iterating in the range 
    # 1 to N
    for i in range(1, n + 1):
  
        summ += Icosidigonal_num(i)
      
    return summ
    
# Driver code 
if __name__ == '__main__'
            
    n = 6
      
  
    print(sum_Icosidigonal_num(n)) 

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C#

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// C# program to find the sum of the 
// first N icosidigonal numbers 
using System;
  
class GFG{ 
          
// Function to find the 
// N-th icosidigonal number 
static int Icosidigonal_num(int n) 
              
    // Formula to calculate 
    // nth icosidigonal number 
    return (20 * n * n - 18 * n) / 2; 
          
// Function to find the sum of the 
// first N icosidigonal number 
static int sum_Icosidigonal_num(int n) 
              
    // Variable to store the sum 
    int summ = 0; 
              
    // Iterating in the range 1 to N 
    for(int i = 1; i < n + 1; i++) 
    
          
       // Finding the sum 
       summ += Icosidigonal_num(i); 
    
    return summ; 
      
// Driver code 
public static void Main(string[] args) 
    int n = 6; 
          
    // Display first Nth 
    // icosidigonal numbers 
    Console.WriteLine(sum_Icosidigonal_num(n)); 
  
// This code is contributed by AnkitRai01

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Output:

721

Time complexity: O(N).

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