Demonstration of Pentagonal Number
It seems a little peculiar to use a class to solve this problem. It's not like there's any need to maintain state. All you're doing, in the end, is making it more difficult to use - instead of:

auto result = pentagonal_number(n);


you have to write:

auto pentagonal_number = PentagonalNum{n};
auto result = num_Of_Dots();


or:

auto result = PentagonalNum{n}.num_Of_Dots();


You could throw out the whole class and just keep the num_Of_Dots() member function, let it take iteration as a parameter, and you'd be done. It would be much easier to use, too.

As for the code itself:

#include &lt;stdio.h&gt;


This is not valid in C++ anymore (though it used to be). You need &lt;cstdio&gt;, or better yet, use actual C++ I/O.

unsigned long long int PentagonalNum::num_Of_Dots()
{
    int result = 1;
    int num = 5;

    for (int i = 1; i &lt; iteration; ++i)
    {
        result += num;
        num += 5;
    }
    return result;
}


It seems like this function can be both noexcept and constexpr (and, if it's going to be a class member, also const). Note that the function requires iteration to be greater than or equal to 1; if you're not checking that here so the function can be noexcept, then you should check it in the constructor.

The big question I have is why result is only an int if you intend to return a unsigned long long. Seems a bit of a waste of all that integer range.

You use some magic numbers and variables with mysterious names (num?). Your code would be clearer and easier to understand with more explicit variable names, like:

constexpr unsigned long long int PentagonalNum(int iteration)
{
    // might want to assert that iteration &gt;= 1

    constexpr auto NUMBER_OF_SIDES = 5uLL;

    auto result = 1uLL; // start with 1 point
    auto points_to_add_on_next_iteration = NUMBER_OF_SIDES;

    for (auto i = 1; i &lt; iteration; ++i)
    {
        result += points_to_add_on_next_iteration;
        points_to_add_on_next_iteration += NUMBER_OF_SIDES;
    }

    return result;
}


As an aside, I suspect the reason this challenge was marked as "hard" is because you are supposed to try and solve it without iterating. In other words, you are supposed to stop and think about the problem and derive an equation for the number of dots at iteration n, rather than just charging in and doing an iterative solution.