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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

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    $\begingroup$ you might want to look at binomial coefficients en.wikipedia.org/wiki/Binomial_coefficient $\endgroup$ – wonko Jul 4 '17 at 15:29
  • $\begingroup$ $$\int\frac{x^2+x}2~\mathrm dx=\frac{x^3+\frac32x^2}6+C$$ $\endgroup$ – Simply Beautiful Art Jul 4 '17 at 15:34
  • $\begingroup$ @SimplyBeautifulArt ... So I was way off. $\endgroup$ – Alexander Day Jul 4 '17 at 15:39
  • $\begingroup$ Except I don't think that's the right answer to the original question, either. $\endgroup$ – Peter Shor Jul 4 '17 at 15:41
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    $\begingroup$ XD @hypergeometric good sigma joke $\endgroup$ – Alexander Day Jul 4 '17 at 16:25
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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: $$1+5+15\cdots+\frac{n(n+1)(n+2)(n+3)}{24}=?$$

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  • $\begingroup$ Multiply by $\frac{n+4}{5}$ $\endgroup$ – Alexander Day Jul 4 '17 at 16:21
  • $\begingroup$ That is right. Luck. $\endgroup$ – farruhota Jul 4 '17 at 16:26
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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

$$N_{d,k}=\binom{k+d-1}d=\frac{k(k+1)\cdots(k+d-1)}{1\cdot2\cdot\cdots d}.$$

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  • $\begingroup$ So the top is $K^3+3K^2+2K$ $\endgroup$ – Alexander Day Jul 4 '17 at 16:12
  • $\begingroup$ @AlexanderDay: say the numerator. $\endgroup$ – Yves Daoust Jul 4 '17 at 16:13
  • $\begingroup$ And would the denominator would be y!, for y dimensions $\endgroup$ – Alexander Day Jul 4 '17 at 16:14
  • $\begingroup$ @AlexanderDay: as written. $\endgroup$ – Yves Daoust Jul 4 '17 at 16:16
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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

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