Trinomial Pascal's Triangle

Question

I know that there's a trinomial theorem (and a multinomial theorem), but I was wondering if there was a similar structure for trinomials as there is for binomials, like Pascal's triangle.

Thanks in advance!

Answer 1

Pascal's pyramid is a series of triangles that tell us the coefficients of terms in powers of trinomials. If the triangles are stacked in succession from top to bottom, we get a pyramid, so it's called Pascal's pyramid. This can be really interesting to explore, so I suggest that you try to calculate the first few triangles.

The way it works is that you start with $1$ around all three corners and then, on the inside, you add up the numbers from the previous triangle to get your new term. The first triangle is: $$1$$ Then we have: $$1 \\ 1 \ \ \ 1$$ Now, the third row is when the new terms come in. This is a three-sided triangle, so we have two middle terms. These middle terms are below the $1$ and $1$ on the edge of the previous triangle, so we get $1+1=2$. $$1 \\ 2 \ \ 2 \\ 1 \ \ 2 \ \ 1$$ Now, the next is a four-sided triangle, so in the middle of the edges, we have two elements. Each of these are below a $1$ and a $2$, so we get $1+2=3$. Also, we have a middle that is below a $2$, a $2$, and a $2$, giving us $2+2+2=6$: $$1 \\ 3 \ \ 3 \\ 3 \ \ 6 \ \ 3 \\ 1 \ \ 3 \ \ 3 \ \ 1$$ From then on, it gets more complicated, but hopefully, you get the point. Can you figure out a formula to connect the elements in each triangle to the trinomial coefficient $\frac{n!}{x!y!x!}$? Can you figure out how to connect each element to the coefficient of a term in the power of a trinomial? Good luck!

Answer 2

Binomial Coefficients

Pascal's Triangle gives the coefficients of the binomial \begin{align*} (1+x)^n\qquad\qquad n\geq 0 \end{align*} we obtain for $n=0,\ldots,4$ \begin{array}{ccccccccccl} &&&&&1&&&&\qquad&\qquad(1+x)^0=1\\ &&&&1&&1&&&\qquad&\qquad(1+x)^1=1+1x\\ &&&1&&\color{blue}{2}&&\color{blue}{1}&&\qquad&\qquad(1+x)^2=1+\color{blue}{2}x+\color{blue}{1}x^2\\ &&1&&3&&\color{blue}{3}&&1&\qquad&\qquad(1+x)^3=1+3x+\color{blue}{3}x^2+1x^3\\ &1&&4&&6&&4&&1\qquad&\qquad(1+x)^4=1+4x+6x^2+4x^3+1x^4 \end{array}

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Let's write the binomial coefficient $\binom{n}{k}$ as $\binom{n}{k,2}$. Using appropriate initial values we have the well-known recurrence relation \begin{align*} \binom{n}{k,2}=\binom{n-1}{k-1,2}+\binom{n-1}{k,2}\qquad\qquad n,k\geq 1 \end{align*} One instance of the recursion is shown in $\rm{\color{blue}{blue}}$ in the triangle above.

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A combinatorial interpretation is based upon lattice paths on a $\mathbb{Z}\times\mathbb{Z}$ grid with steps $(1,1),$ and $(1,-1)$. Walks of length $n$ can be represented by the generating function \begin{align*} \left(xy+xy^{-1}\right)^n\qquad\text{or}\qquad\left(1+x\right)^n \end{align*}

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Trinomial Coefficients

A corresponding construction with trinomial coefficients is based upon the trinomial \begin{align*} (1+x+x^2)^n\qquad\qquad n\geq 0 \end{align*} we obtain for $n=0,\ldots,4$

\begin{array}{ccccccccccl} &&&&1&&&&&&(1+x+x^2)^0 =1\\ &&&1&1&1&&&&&(1+x+x^2)^1 =1+1x+1x^2\\ &&1&2&\color{blue}{3}&\color{blue}{2}&\color{blue}{1}&&&&(1+x+x^2)^2 =1+2x+\color{blue}{3}x^2+\color{blue}{2}x^3+\color{blue}{1}x^4\\ &1&3&6&7&\color{blue}{6}&3&1&&&(1+x+x^2)^3 =1+3x+6x^2+7x^3+\color{blue}{6}x^4+3x^5+1x^6\\ 1&4&10&16&19&16&10&4&1&&(1+x+x^2)^4 =1+4x+10x^2+16x^3+19x^4\\ &&&&&&&&&&\qquad\qquad\qquad\qquad\qquad+16x^5+10x^6+4x^7+1x^8 \end{array}

Let's write the trinomial coefficient as $\binom{n}{k,3}$. Using appropriate initial values we have the recurrence relation \begin{align*} \binom{n}{k,3}=\binom{n-1}{k-1,3}+\binom{n-1}{k,3}+\binom{n-1}{k+1,3}\qquad\qquad n,k\geq 1 \end{align*} One instance of the recursion is shown in $\rm{\color{blue}{blue}}$ in the triangle above.

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A combinatorial interpretation is based upon lattice paths on a $\mathbb{Z}\times\mathbb{Z}$ grid with steps $(1,1),(1,0)$ and $(1,-1)$. Walks of length $n$ can be represented by the generating function \begin{align*} \left(xy+x+xy^{-1}\right)^n\qquad\text{or}\qquad\left(1+x+x^2\right)^n \end{align*}