Find the sum of the first N Dodecagonal Numbers
Last Updated :
23 Nov, 2022
Given a number N the task is to find the sum of first N Dodecagonal Number.
The first few dodecagonal numbers are 1, 12, 33, 64, 105, 156, 217 …
Examples:
Input: N = 3
Output: 46
Explanation:
1, 12 and 33 are the first three Dodecagonal numbers
Input: N = 5
Output: 215
Approach:
- Initially, we need to create a function that will help us to calculate the Nth Dodecagonal number.
- Run a loop starting from 1 to N, to find ith Dodecagonal number.
- Add all the above calculated Dodecagonal numbers.
- Finally, display the sum of the first N Dodecagonal numbers.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int Dodecagonal_num( int n)
{
return (5 * n * n - 4 * n);
}
int sum_Dodecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += Dodecagonal_num(i);
}
return summ;
}
int main()
{
int n = 5;
cout << (sum_Dodecagonal_num(n));
return 0;
}
|
Java
class GFG {
static int Dodecagonal_num( int n)
{
return ( 5 * n * n - 4 * n);
}
static int sum_Dodecagonal_num( int n)
{
int summ = 0 ;
for ( int i = 1 ; i < n + 1 ; i++)
{
summ += Dodecagonal_num(i);
}
return summ;
}
public static void main(String[] args)
{
int n = 5 ;
System.out.println(sum_Dodecagonal_num(n));
}
}
|
Python3
def Dodecagonal_num(n):
return ( 5 * n * n - 4 * n)
def sum_Dodecagonal_num(n) :
summ = 0
for i in range ( 1 , n + 1 ):
summ + = Dodecagonal_num(i)
return summ
if __name__ = = '__main__' :
n = 5
print (sum_Dodecagonal_num(n))
|
C#
using System;
class GFG {
static int Dodecagonal_num( int n)
{
return (5 * n * n - 4 * n);
}
static int sum_Dodecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += Dodecagonal_num(i);
}
return summ;
}
public static void Main(String[] args)
{
int n = 5;
Console.WriteLine(sum_Dodecagonal_num(n));
}
}
|
Javascript
<script>
function Dodecagonal_num(n)
{
return (5 * n * n - 4 * n);
}
function sum_Dodecagonal_num(n)
{
let summ = 0;
for (let i = 1; i < n + 1; i++)
{
summ += Dodecagonal_num(i);
}
return summ;
}
let n = 5;
document.write(sum_Dodecagonal_num(n));
</script>
|
Time Complexity: O(N).
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...