Find the sum of the first Nth Heptadecagonal Number
Given a number N, the task is to find the sum of first N Heptadecagonal Numbers.
The first few heptadecagonal numbers are 1, 17, 48, 94, 155, 231 …
Examples:
Input: N = 3
Output: 66
Explanation:
1, 17 and 48 are the first three heptadecagonal numbers.
Input: N = 6
Output: 546
Approach:
- Initially, we need to create a function which will help us to calculate the Nth heptadecagonal number.
- Now, run a loop starting from 1 to N, to find ith heptadecagonal number.
- Add all the above calculated heptadecagonal numbers.
- Finally, display the sum of the first N heptadecagonal numbers.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int heptadecagonal_num( int n)
{
return ((15 * n * n) - 13 * n) / 2;
}
int sum_heptadecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += heptadecagonal_num(i);
}
return summ;
}
int main()
{
int n = 5;
cout << sum_heptadecagonal_num(n);
}
|
Java
class GFG{
public static int heptadecagonal_num( int n)
{
return (( 15 * n * n) - 13 * n) / 2 ;
}
public static int sum_heptadecagonal_num( int n)
{
int summ = 0 ;
for ( int i = 1 ; i < n + 1 ; i++)
{
summ += heptadecagonal_num(i);
}
return summ;
}
public static void main(String[] args)
{
int n = 5 ;
System.out.println(sum_heptadecagonal_num(n));
}
}
|
Python3
def heptadecagonal_num(n):
return (( 15 * n * n) - 13 * n) / / 2
def sum_heptadecagonal_num(n) :
summ = 0
for i in range ( 1 , n + 1 ):
summ + = heptadecagonal_num(i)
return summ
if __name__ = = '__main__' :
n = 5
print (sum_heptadecagonal_num(n))
|
C#
using System;
class GFG{
public static int heptadecagonal_num( int n)
{
return ((15 * n * n) - 13 * n) / 2;
}
public static int sum_heptadecagonal_num( int n)
{
int summ = 0;
for ( int i = 1; i < n + 1; i++)
{
summ += heptadecagonal_num(i);
}
return summ;
}
public static void Main()
{
int n = 5;
Console.WriteLine(sum_heptadecagonal_num(n));
}
}
|
Javascript
<script>
function heptadecagonal_num(n)
{
return ((15 * n * n) - 13 * n) / 2;
}
function sum_heptadecagonal_num(n)
{
let summ = 0;
for (let i = 1; i < n + 1; i++)
{
summ += heptadecagonal_num(i);
}
return summ;
}
let n = 5;
document.write(sum_heptadecagonal_num(n));
</script>
|
Time complexity: O(N).
Auxiliary space: O(1) since it is using constant space for variables
Last Updated :
19 Sep, 2022
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