# The relationship between the Pascal's triangle/sequence and the binomial number theorem?

What is the relationship between Pascal's sequence and the binomial theorem? I want to have a thorough and intuitive understanding of the connections between the two.

Though I am able to relate to the fact that:

$(x+y)^n = \left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)x^{n-1}y+\left(\begin{array}{c}n\\ 2\end{array}\right)x^{n-2}y^2+.....+\left(\begin{array}{c}n\\ n\end{array}\right)y^n$

And $\left(\begin{array}{c}n\\ 0\end{array}\right)$ is the first element from the Pascal sequence. But can someone help me by giving an intuitive description of the relationship between the two? I want to be able to thoroughly understand the connections.

• What exactly is it that you call the "triangular number sequence"? Is it not $1, 3, 6, 10, 15, 21,\ldots$? – Arthur Jun 28 '18 at 7:18
• I’m not sure I understand the question. It looks like you’ve just said “the first number in the triangular number sequence is 1 and so is the first term in any binomial expansion”. – Benjamin Jun 28 '18 at 7:19
• The triangular numbers are ${n \choose 2}$ for $n \ge 1$, thus the coefficient of $x^{n-2} y^2$ in $(x+y)^n$. – Robert Israel Jun 28 '18 at 7:23
• I'm really sorry. I, for a moment, confused the triangular number sequence with Pascal's triangle. Very sorry for the inconveniences. I typed it wrong. – Ramana Jun 28 '18 at 7:47

To get from the $n$th to the $(n+1)$st row, you add adjacent coefficients. This corresponds to the identity ${n+1\choose k}={n\choose k}+{n\choose k-1}$.

Intuitively, this identity just says that if you take one element out, you can count $k$ element subsets with and without that element, then combine to get all $k$ element subsets...

Think
$$(x+y)^n = (x+y) (x+y) \cdots (x+y)$$ with $n$ factors on the right. When you expand this using the distributive law there will be $2^n$ terms, corresponding to the ways to choose either $x$ or $y$ from each of the factors.

When you collect the terms there will be just $\binom{n}{k}$ ways to have chosen the $x$ just $k$ times. That will be the coefficient of $x^ky^{n-k}$.

Let's expand, for instance, $$(l+r)^5$$. Expanding consists in choosing either $$l$$ or $$r$$ in each of the 5 brackets, which is equivalent to choosing either left or right in each level of the lattice below:

The coefficient of each factor ($$l^5$$, $$l^4r$$...) equals the number of paths that go from the peak of the lattice to the bottom node corresponding to that factor.

We need to count the number of paths, from the peak to the bottom nodes. The two top levels are quite clear,

You can get to the $$?$$ from the green node or from the red node. Since there are exactly two paths from the peak to the green node and one path from the peak to the red node, we have a total of $$2+1=3$$ paths from the peak to the $$?$$. That is, the number of paths from the peak to each node is exactly the sum of the paths ending at the two nodes above. But that is how one builts the Pascal's triangle.