Tetrahedral numbers: what is the equation omitting the $\Sigma$ 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
 A: By a smart generalization of the pattern, you can attempt the following formula for the tetrahedral numbers:
$$T_k=\frac{k(k+1)(k+2)}{1\cdot2\cdot3}.$$
Then you test the hypothesis by using the fact that the difference between two successive tetrahedral numbers, is a triangular number:
$$T_k-T_{k-1}=\frac{k(k+1)}{1\cdot2}.$$
We indeed have
$$T_k-T_{k-1}=\frac{k(k+1)(k+2)-(k-1)k(k+1)}{1\cdot2\cdot3}=\frac{3k(k+1)}{1\cdot2\cdot3}.$$
You will understand that this generalizes to higher order ($d$ dimensions), giving the formula 
$$N_{d,k}=\binom{k+d-1}d=\frac{k(k+1)\cdots(k+d-1)}{1\cdot2\cdot\cdots d}.$$
A: Note:
$$\underbrace{1+1+1+\cdots+1}_{n}=n$$
$$1+2+3+\cdots+n=\frac{n(n+1)}{2}$$
$$1+3+6+\cdots+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}$$
$$1+4+10+\cdots+\frac{n(n+1)(n+2)}{6}=\frac{n(n+1)(n+2)(n+3)}{24}$$
Can you guess the sum formula:
$$1+5+15\cdots+\frac{n(n+1)(n+2)(n+3)}{24}=?$$
A: Note that 
$$\begin{align}
&\sum_{r=1}^n \binom r1&=\binom {n+1}2&=\frac {n(n+1)}2\\
&\sum_{r=1}^n \binom {r+1}2&=\binom {n+2}3&=\frac {n(n+1)(n+2)}6\\
&\sum_{r=1}^n \binom {r+1}3&=\binom {n+3}4&=\frac {n(n+1)(n+2)(n+3)}{24}\\
&&\vdots&\\
&\sum_{r=1}^n \binom {r+1}m&=\binom {n+m}{m+1}&=\frac {n^{\overline{m+1}}}{m+1}\end{align}$$
Perhaps this is what you are looking for.

Useful references:
Binomial coefficients
Hockey-stick identity
Rising factorials
