0

The following are the triangular numbers. 

rank = 1   2   3  4   5   6        
term = 1   3   6  10  15  21

A rule for triangular numbers is: r(r + 1) / 2
I tried to derive that with Pascal's trangle (bellow) but I ended up with (r + 1)(r + 2)/2 instead. 

1   3   6  10   15   21  
  2   3   4    5    6  
    1   1   1    1  
      0   0   0  
        0   0  
          0

I could derive r(r + 1)/2 if I added rank and term 0. I.e. 

rank = 0  1   2   3  4   5   6        
term = 0  1   3   6  10  15  21

But why is this? I think 0 is not a triangular number. So how does this work? Does the pascal's triangle has a prerequisite to start from 0?

Jim
  • 727
  • Using Pascal's triangle, did you try to evaluate ${n+1}\choose{n-1}$? – user328442 Jun 14 '16 at 18:13
  • @user328442:I don't understand what you mean. Sorry – Jim Jun 14 '16 at 18:14
  • Why downvote?Is it a bad question? Can I improve the post? – Jim Jun 14 '16 at 18:14
  • In Pascal's triangle, it is common to use "choose" notation. You seem to be looking for the sequence defined by ${n+1}\choose{n-1}$. An example of how this notation works is the following: $n\choose{k}$ = $\frac{n!}{k!(n-k)!}$. So, instead of using ${n+2}\choose{n}$, which leads to what you ended up with, simply rewrite your sequence with ${n+1}\choose{n-1}$ to fix the problem. – user328442 Jun 14 '16 at 18:19
  • As $0=\frac {0 \cdot 1}2$ it fits your formula for a triangular number. That makes sense, a triangle of side $0$ has $0$ items in it. – Ross Millikan Jun 14 '16 at 19:44
  • @RossMillikan:But usually 0 is ommited. Why would I count 0 then as part of the series e.g. https://upload.wikimedia.org/wikipedia/commons/thumb/1/1c/First_six_triangular_numbers.svg/374px-First_six_triangular_numbers.svg.png – Jim Jun 14 '16 at 20:21

2 Answers2

1

The triangular numbers are

$$T_n:=\sum_{k=0}^nk=\frac{n(n+1)}2=\binom n2.$$ They appear as a diagonal row in Pascal's triangle, which starts at $\color{blue}{n=0,r=0}$, with$\color{blue}{\binom00=1}$.

$$\color{blue}1\\1\ \ \ 1\\1\ \ \ 2\ \ \ \color{green}1\\1\ \ \ 3\ \ \ \color{green}3\ \ \ 1\\1\ \ \ 4\ \ \ \color{green}6\ \ \ 4\ \ \ 1\\1\ \ \ 5\ \ \ \color{green}{10}\ \ \ 10\ \ \ 5\ \ \ 1$$

By convention, all elements outside the triangle are zero. This is compatible with

$$T_0=\frac{0\cdot1}2=\binom12=0,\\T_1=\frac{1\cdot2}2=\binom22=\color{green}1,\\T_2=\frac{2\cdot3}2=\binom32=\color{green}3,\\T_3=\frac{3\cdot4}2=\binom42=\color{green}6\\\cdots$$

0

Your problem has a question. Regardless of whether you add a $0$-th term with value $0$, every other $k$-th term sill has the same value, namely $\frac{k(k+1)}{2}$. So I do not see why you say you have "derived something else with Pascal's triangle", because (as far as we know) it's just not possible to derive something wrong that way. You're probably misusing whatever it is you used (which you didn't specify so we don't know).

If you look at the Pascal's triangle (with the apex at the top), the triangular numbers form the $2$nd diagonal, with the $0$-th one being the leftmost diagonal with only ones. The $k$-th number in the $2$nd diagonal is the binomial coeficient $C(k+1,2) = \frac{k(k+1)}{2}$, so there's nothing wrong with Pascal's triangle.

user21820
  • 60,745