Find the sum of the first N Centered Dodecagonal Number
Last Updated :
30 Jan, 2023
Given a number N, the task is to find the sum of first N Centered Dodecagonal Number.
The first few Centered Dodecagonal Numbers are 1, 13, 37, 73, 121, 181 …
Examples:
Input: N = 3
Output: 51
Explanation:
1, 13 and 37 are the first three centered Dodecagonal number.
Input: N = 5
Output: 245
Approach:
- Initially, create a function which will help us to calculate the Nth Centered Dodecagonal number.
- Run a loop starting from 1 to N, to find i-th Centered Dodecagonal number.
- Add all the above calculated Centered Dodecagonal numbers.
- Finally, display the sum of the first N Centered Dodecagonal numbers.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int Centered_Dodecagonal_num( int n)
{
return 6 * n * (n - 1) + 1;
}
int sum_Centered_Dodecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += Centered_Dodecagonal_num(i);
}
return summ;
}
int main()
{
int n = 5;
cout << sum_Centered_Dodecagonal_num(n);
}
|
Java
class GFG {
static int Centered_Dodecagonal_num( int n)
{
return 6 * n * (n - 1 ) + 1 ;
}
static int sum_Centered_Dodecagonal_num( int n)
{
int summ = 0 ;
for ( int i = 1 ; i < n + 1 ; i++)
{
summ += Centered_Dodecagonal_num(i);
}
return summ;
}
public static void main (String[] args)
{
int n = 5 ;
System.out.print(sum_Centered_Dodecagonal_num(n));
}
}
|
Python3
def Centered_Dodecagonal_num(n):
return 6 * n * (n - 1 ) + 1
def sum_Centered_Dodecagonal_num(n) :
summ = 0
for i in range ( 1 , n + 1 ):
summ + = Centered_Dodecagonal_num(i)
return summ
if __name__ = = '__main__' :
n = 5
print (sum_Centered_Dodecagonal_num(n))
|
C#
using System;
class GFG{
static int Centered_Dodecagonal_num( int n)
{
return 6 * n * (n - 1) + 1;
}
static int sum_Centered_Dodecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += Centered_Dodecagonal_num(i);
}
return summ;
}
public static void Main()
{
int n = 5;
Console.Write(sum_Centered_Dodecagonal_num(n));
}
}
|
Javascript
<script>
function Centered_Dodecagonal_num(n)
{
return 6 * n * (n - 1) + 1;
}
function sum_Centered_Dodecagonal_num(n)
{
let summ = 0;
for (let i = 1; i < n + 1; i++)
{
summ += Centered_Dodecagonal_num(i);
}
return summ;
}
let n = 5;
document.write(sum_Centered_Dodecagonal_num(n));
</script>
|
Time Complexity: O(N).
Auxiliary Space: O(1) since constant variables are being used
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