# Pascal's Triangle

**Pascal's triangle** is a triangular array constructed by summing adjacent elements in preceding rows. Pascal's triangle contains the values of the binomial coefficient. It is named after the $17^\text{th}$ century French mathematician, Blaise Pascal (1623 - 1662).

$1\\ 1\quad 1\\ 1\quad 2 \quad 1\\ 1\quad 3 \quad 3 \quad 1\\ 1\quad 4 \quad 6 \quad 4 \quad 1\\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ \cdots$

Pascal's triangle can be used to visualize many properties of the binomial coefficient and the binomial theorem.

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## Construction of Pascal's Triangle

Begin by placing a $1$ at the top center of a piece of paper. The next row down of the triangle is constructed by summing adjacent elements in the previous row. Because there is nothing next to the $1$ in the top row, the adjacent elements are considered to be $0:$

This process is repeated to produce each subsequent row:

This can be repeated indefinitely; Pascal's triangle has an infinite number of rows:

## Notation of Pascal's Triangle

The topmost row in Pascal's triangle is considered to be the $0^\text{th}$ row. The next row down is the $1^\text{st}$ row, then the $2^\text{nd}$ row, and so on. The leftmost element in each row is considered to be the $0^\text{th}$ element in that row. Then, to the right of that element is the $1^\text{st}$ element in that row, then the $2^\text{nd}$ element in that row, and so on.

With this convention, each $i^\text{th}$ row in Pascal's triangle contains $i+1$ elements.

The convention of beginning the order with $0$ may seem strange, but this is done so that the elements in the array correspond to the values of the binomial coefficient.

Let $x_{i,j}$ be the $j^\text{th}$ element in the $i^\text{th}$ row of Pascal's triangle, with $0\le j\le i$. Then,

$x_{i,j}=\binom{i}{j}.$

This property of Pascal's triangle is a consequence of how it is constructed and the following identity:

Let $n$ and $k$ be integers such that $1\le k\le n$. Then,

$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}.$

Using Pascal's triangle, what is $\binom{6}{2}$?

Look for the $2^\text{nd}$ element in the $6^\text{th}$ row. The value of that element will be $\binom{6}{2}$.

Thus, $\binom{6}{2}=15$. $_\square$

$\begin{array}{rc} 0^\text{th} \text{ row:} & 1 \\ 1^\text{st} \text{ row:} & 1 \quad 1 \\ 2^\text{nd} \text{ row:} & 1 \quad 2 \quad 1 \\ 3^\text{rd} \text{ row:} & 1 \quad 3 \quad 3 \quad 1 \\ 4^\text{th} \text{ row:} & 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ \vdots \ \ \ & \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \quad \cdot \end{array}$

Pascal's triangle is shown above for the $0^\text{th}$ row through the $4^\text{th}$ row. What is the $4^\text{th}$ element in the $10^\text{th}$ row?

$$

**Note:** Each row starts with the $0^\text{th}$ element. For example, the $0^\text{th}$, $1^\text{st}$, $2^\text{nd}$, and $3^\text{rd}$ elements of the $3^\text{rd}$ row are 1, 3, 3, and 1, respectively.

## Patterns in Pascal's Triangle

Pascal's triangle is a way to visualize many patterns involving the binomial coefficient. Here are some of the ways this can be done:

The $n^\text{th}$ row of Pascal's triangle contains the coefficients of the expanded polynomial $(x+y)^n$.

Expand $(x+y)^4$ using Pascal's triangle.

The $4^\text{th}$ row will contain the coefficients of the expanded polynomial.

$(x+y)^4=\color{blue}{1}x^4+\color{blue}{4}x^3y+\color{blue}{6}x^2y^2+\color{blue}{4}xy^3+\color{blue}{1}y^4$

Start at any of the "$1$" elements on the left or right side of Pascal's triangle. Sum elements diagonally in a straight line, and stop at any time. Then, the next element down diagonally in the opposite direction will equal that sum.

If you start at the $r^\text{th}$ row and end on the $n^\text{th}$ row, this sum is

$\sum\limits_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}.$

Using Pascal's triangle, what is $\displaystyle\sum\limits_{k=2}^{5}\binom{k}{2}?$

Start at a $1$ on the $2^\text{nd}$ row, and sum elements diagonally in a straight line until the $5^\text{th}$ row:

$1+3+6+10=20.$

Or, simply look at the next element down diagonally in the opposite direction, which is $20$. $_\square$

$\begin{array}{c} 1 \end{array} \\ \begin{array}{cc} 1 & 1 \end{array} \\ \begin{array}{ccc} 1 & 2 & \color{red}{1}\end{array} \\ \begin{array}{cccc} 1 & 3 & \color{red}{3} & 1\end{array} \\ \begin{array}{ccccc} 1 & 4 & \color{red}{6} & 4 & 1\end{array} \\ \begin{array}{cccccc} \vdots & \hphantom{\vdots} & \vdots & \hphantom{\vdots} & \vdots \end{array}\\ \begin{array}{cccccc} 1 & 25 & \color{red}{300} & 2300 & 12650 & \cdots \end{array} \\ \begin{array}{ccccccc} 1 & 26 & 325 & 2600 & 14950 & \cdots & \hphantom{1000} \end{array} \\$

Pascal's triangle is shown above for the $0^\text{th}$ row through the $4^\text{th}$ row, and parts of the $25^\text{th}$ and $26^\text{th}$ rows are also shown above.

What is the sum of all the $2^\text{nd}$ elements of each row up to the $25^\text{th}$ row?

$$

**Note**: The visible elements to be summed are highlighted in red.

**Additional clarification**: The topmost row in Pascal's triangle is the $0^\text{th}$ row. Then, the next row down is the $1^\text{st}$ row, and so on. The leftmost element in each row of Pascal's triangle is the $0^\text{th}$ element. Then, the element to the right of that is the $1^\text{st}$ element in that row, and so on.

The sum of the elements in the $n^\text{th}$ row of Pascal's triangle is equal to $2^n$.

This is a way to express the identity

$\sum\limits_{k=0}^{n}\binom{n}{k}=2^n.$

$\begin{array}{c} 1 \end{array} \\ \begin{array}{cc} 1 & 1 \end{array} \\ \begin{array}{ccc} 1 & 2 & 1 \end{array} \\ \begin{array}{cccc} 1 & 3 & 3 & 1\end{array} \\ \begin{array}{ccccc} 1 & 4 & 6 & 4 & 1\end{array} \\ \begin{array}{cccccc} \vdots & \hphantom{\vdots} & \vdots & \hphantom{\vdots} & \vdots \end{array} \\$

Pascal's triangle is shown above for the $0^\text{th}$ row through the $4^\text{th}$ row.

What is the sum of all the elements in the $12^\text{th}$ row?

$$

**Note**: The topmost row in Pascal's triangle is the $0^\text{th}$ row. Then, the next row down is the $1^\text{st}$ row, and so on.

The following property follows directly from the hockey stick identity above:

The $2^\text{nd}$ element in the $(n+1)^\text{th}$ row is the $n^\text{th}$ triangular number.

This is a way to express the identity

$\sum\limits_{k=1}^{n}{k}=\binom{n+1}{2}.$

Construct a Pascal's triangle, and shade in even elements and odd elements with different colors. The shading will be in the same pattern as the Sierpinski Gasket:

This is an application of Lucas's theorem.

**Cite as:**Pascal's Triangle.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/pascals-triangle/