0
$\begingroup$

Preface: For images with r and x, r=k and x=n

Recently, I have started looking closer at Pascal's Triangle (because it's fascinating) and I wanted to try to make a general form for the function of each diagonal. Essentially, the diagonals are represented by k, where k=1 refers to the first diagonal which is only filled with ones, k=2 is the second diagonal which is filled with the natural numbers.

The first diagonal would be k=1, where the function would be n. Then, k=2 would give us (n^2)/2 + Cn . I used integration to give me k=2 because k=1 is the rate of change of k=2, so integration k=1 should yield k=2. However, when integrating k=1, the integral should not have an n multiplied by the integration constant, it should be a lone +C. This I am confused on.

I learned that integration works even less as k increases, as shown here where I try to integrate k=2. The function is slightly inaccurate and also requires a random bonus n to be attached to the integration constant. Why doesn't integrating work?

I have recognized that the lowest order term has a pattern to it. This leads me to believe there is an answer, I'm just not sure how to reach it.

My goal is to be able to determine the function of the nth term at a certain value of k, which I decided to label a.

Edit: Fixed x->n and r->k

Edit: Here is an image that can help clarify the process I am going through and how I used integration.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ It is not at all clear what functions you are trying for. You introduce $r$ in a way that is confusing, introduce "$x$" without indicating what it is at all, talk about integration without indicating what the integration is supposed to be accomplishing and why. The coefficients in Pascal's triangle are given by $${n\choose k} = \frac {n!}{k!(n-k)!}$$ The diagonals are what you get when you hold $k$ (or $n-k$) constant while varying $n$. How is what you are after different from that? $\endgroup$ – Paul Sinclair Jun 23 '19 at 15:46
  • $\begingroup$ I edited the post so it is much more coherent. $\endgroup$ – Divide Me By Zero Jun 24 '19 at 1:18
  • $\begingroup$ The diagonals of the triangle are sums of finite numbers of integers. Integrals are designed to solve a very different kind of problem. What made you consider integrals as a possible way to compute the diagonals? $\endgroup$ – David K Jun 24 '19 at 1:56
0
$\begingroup$

You have made some interesting observations on the structure of Pascal's triangle.

First, you should note that the diagonals you are referring to are actually defined by $f(k=a)=\binom{n}{a}$; that is, what you call $f(k=1)$ (for example) is actually $\binom{n}{1}$ for each $n$. Now, the hockey stick identity will tell you that $$\sum_{i=r}^n \binom{i}{r} = \binom{n+1}{r+1},$$ so that the values of $f(k=a)$ are the sequential sums of $f(k=a-1)$. Thus for example for $f(k=5)$ we get $6=1+5$, $21=1+5+15$, $56 = 1+5+15+35$.

So there are a couple of ways you might think about finding a concrete formula. First, of course, the binomial coefficients have a reasonably explicit form: $$\binom{n}{k} = \frac{n\cdot(n-1)\cdots (n-k+1)}{k!}.$$ For small values of $k$, you can write this down explicitly, as you have; of course, the degrees of the result as a polynomial in $n$ gets larger and larger. You might also think about using what you (may) know about sums of powers. For example, to derive $f(k=3)$ you could do \begin{align*} f(k=3)(n) &= \binom{n}{3} = \sum_{i=2}^n\binom{i}{2} \\ &= \sum_{i=2}^n \left(\frac{i^2}{2} - \frac{i}{2}\right) \\ &= \frac{1}{2}\left(\sum_{i=1}^{n-1} i^2 - \sum_{i=2}^{n-1} i\right) \\ &= \frac{1}{2}\left(\frac{(n-1)(n)(2n-1)}{6} - \frac{(n-1)(n)}{2}\right) \\ &= \frac{1}{2}\left(\frac{2n^3 - 3n^2+n}{6} - \frac{n^2-n}{2}\right) \\ &= \frac{1}{2}\cdot\frac{2n^3 - 6n^2 + 4n}{6} \\ &= \frac{n^3}{6} - \frac{n^2}{2} + \frac{n}{3}, \end{align*} which agrees with your formula. You will not succeed in finding a simple closed formula in terms of $n$ and $k$ however.

If you want to look more into representations of sums of powers of consecutive integers, which is obviously related to your question, you could look here or in many other places.

Hope that helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Awesome, thank you! $\endgroup$ – Divide Me By Zero Jun 24 '19 at 2:56

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.