Conjecture + Calculus = Conjectulus?

So here it goes the solution. I'm gonna try to be succinct.

First thing, and the most important thing of the answer (almost 50%) is to admit that the pyramid grown in some pattern (I hope this is the right word). We can make some list, but I'll do it while I write the solution. With that, we can assume our conjecture in a easy way.

So, let's make our small tower's formula. This is simple:

On the first floor, we have 1 tower, on the second floor we have 2 towers, on the third floor, we have 3 towers, and this goes as far as you want to. If we sum the towers of the floors, we will get something like this:

$(1 + 2 + 3 + 4 + 5 + .......... + n)$ . Noticed that the number of towers in determined floor is the same of the floor's number if we count it from the top to the bottom (base of the pyramid).

This can be expressed as: $\sum_{t=1}^nt$ where $n$ is the number of floors and $t$ is the number of towers on determined ( $n$ º) floor.

And this summation can be expressed as: $t = \frac{n(n+1)}{2} = \frac{n^{2} + a}{2}$ . This is our first formula. In order to remember which formula we will find the derived, call this $\textbf A$ .

The second one is about the number of cards that is necessary to build the pyramid. There's something very important to be noticed:

Each floor has shapes of $t$ small triangles pointing upwards, but the base don't have this kind of triangles. Instead, the base has $n$ towers where each "semi-triangles" is missing one side. Look at this image:

And the last information that we need is that every triangle has 3 sides, or in this case, 3 cards.

So, agree with me: If we find the number of small towers, multiply by 3 (number of cards), and sum the difference of the cards that is missing on the base (negative value), we will have our "number of cards formula".

Let "c" be the number of cards:

$c = 3 \times \frac{n^{2} + n}{2} - n \Rightarrow c = \frac{3n^2 + n}{2}$ . Call this $\textbf B$

This is our second formula.

Now comes the third part, the shapes of small triangles. I'm considering the way I drew in the paint so this can be more interesting.

Now I think that the logic of these conjectures is getting more easy. Let's divided this part in two: The triangles that is pointing upwards, and other triangles that is pointing down.

Use this image in the case of some doubt shows up:

First floor, 1 triangle, second floor, 2 triangles, third floor, 3 triangles... until the base, which has no upwards triangles. The base has $n$ small towers, so: if we find the number of towers and sum the difference of towers that it's not a triangle, we will find the number of triangles pointing upwards. Call this " $k$ "

$\ k = \frac{n^{2} + n}{2} - n \Rightarrow k = \frac{n^{2} - n}{2}$

The logic for the triangle that are pointing down is the following: they only start to shows up on the second floor (if there's one), and in the base they are triangles too. So, the only thing to do is to adapt our last formula to $n-1$ and no need to sum some difference. You can also look to the pyramid and notice that the number of triangles pointing upwards is the same of pointing down. But let's make our algebra: Call this " $l$ "

$l = \frac{(n-1)^2 + (n-1)}{2} \Rightarrow l = \frac{n^{2} - n}{2}$

Now, we already know the number of triangles, just sum the formulas (which are the same), and we have: Let number of triangles $(t$ ).

$t = n^{2} - n$ . Call this $\textbf C$ .

So there it is our 3 formulas!!!

Here Mohamed imagine this pyramid as a tree (upside down, obviously).Now we calculate the derivative (which I'm gonna jump the step of explaining it because I already write a lot).

$\frac{\mathrm{d} }{\mathrm{d} n}A = \frac{2n + 1}{2}$

$\frac{\mathrm{d} }{\mathrm{d} n}B = \frac{6n + 1}{2}$

$\frac{\mathrm{d} }{\mathrm{d} n}C = 2n - 1$

Put in the place of $n$ the value $68$ and make the sum. The result is $408$ .

I hope I made it clear.