I have heard of the famous Johann Carl Friedrich Gauss, and I've also heard of his triangular formula. Which goes like this:
When Gauss was young, his teacher gave him and the rest of his class the job of adding all the numbers from $1$ to $100$. The teacher was trying to get some time to herself, but this turned out to be futile. Gauss was able to solve it in some $15$ seconds, using a formula he came up with. Instead of using this: $$ \sum^{100}_1 x=y $$ He found a cunning technique to solve it. He noticed that if you broke the sequence in half and put the two ends above each other, like so: $$ \begin{matrix} 100&99&98&97&96&95&\cdots&51\\ 1&2&3&4&5&6&\cdots&50\\ \end{matrix} $$ If you notice the columns, you will see that they add up to $101$.since there are $50$ of these, you get one simple equation, which is this: $$101\times 50=5050$$ After surprising his teacher, he replaced some of the numbers with $x$, to get $$(x+1) \cdot \frac{x}{2}=\triangle x$$
(Yes, i know I'm not using the triangle correctly) anyway, I have also read that the other famous triangle, Pascal's triangle, has an unusual property; the first layer on the left is full of $1$'s, the next is the counting numbers, then the triangular numbers, and then the tetrahedral numbers. I then came up with the idea that if there is a formula for triangular numbers, then there should be a tetrahedral formula. To begin, I used the sigma function: $$\sum^x_1 (x+1) \cdot \frac{x}{2}=\triangle \triangle x$$ (Again, I didn't use the triangle sign correctly, and two stand for tetrahedron) To begin my idea, I simplify Gauss's formula into a quadratic formula, which is $$\frac{x^2+x}{2}$$ This is the same as the original formula, just simplified. To then get an idea of calculating tetrahedral numbers. I took the triangular numbers to get the $8$th tetrahedral number, which are: $$\begin{matrix} 1&3&6&10&15&21&28&36 \end{matrix}$$ Adding the ends gets $$\frac{x^2+x}{2}+1$$ With there being $\frac{x}{2}$ of these, I redid the formula again. $$\left(\frac{x^2+x}{2}+1\right)\cdot \frac{x}{2}=\frac{x^3+x^2+2x}{4}$$ But this formula is still missing something, a number $y$, which is some number that directly relates to x somehow. Because when I put this formula through the first few numbers; x is number, z is formula number, c is correct number $$\begin{align} x=1&\rightarrow z=1 \text{ and }c=1\\ x=2&\rightarrow z=4 \text{ and }c=4\\ x=3&\rightarrow z=10.5 \text{ but }c=10\\ x=4&\rightarrow z=22 \text{ but }c=20\\ x=5&\rightarrow z=40 \text{ but }c=35\\ x=6&\rightarrow z=54 \text{ but }c=56\\ x=7&\rightarrow z=101.5 \text{ but }c=84\\ \end{align}$$ As you can see,this is wrong. The first two are accurate, but the rest are wrong, and I don't get what happened with $6$.I don't think this is the correct formula, so I'll use differential and integral calculus to solve this. If $f(x)=\triangle \triangle x$, then $f'(x)=\triangle x$ and
$$\int \frac{x^2+x}{2}=y$$
Unfortunately, I am bad at integrals. Could someone show me how to integrate this? Thanks in advance