2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ XD @hypergeometric good sigma joke $\endgroup$ – Alexander Day Jul 4 '17 at 16:25
0
$\begingroup$

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: $$1+5+15\cdots+\frac{n(n+1)(n+2)(n+3)}{24}=?$$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.