## Introduction to Pascal’s triangle

One of the most interesting number pattern in mathematics is **Pascal’s triangle**. This pattern is developed by a famous french mathematician and philosopher,
**Blaise Pascal**.

To build the Pascal’s triangle start with 1 at top, then continue placing numbers below it in a triangular pattern.

The two side edges always have 1s placed on them.

Each number in Pascal’s triangle is the sum of two numbers just around it in the above row.

In above figure, which is a Pascal’s triangle,

There are only 1s numbers placed on the two edges.

In 3^{rd} row, **2 is sum of 1 and 1** placed in 2^{nd} row.

In 4^{th} row, **3 is sum 1 and 2** placed in 3^{rd} row.

In 5^{th} row, **4 is sum 1 and 3** and **6 is sum of 3 and 3** placed in 4^{th} row.

In 6^{th} row, **6 is sum 1 and 5**, **15 is sum of 5 and 10** and **20 is sum of 10 and
10** placed in 5^{th} row.

In 7^{th} row, **7 is sum 1 and 6**, **21 is sum of 6 and 15** and **35 is sum of 20 and
15** placed in 6^{th} row.

## Number pattern on diagonals of Pascal’s triangle

The above figure, shows an interesting pattern formed with the numbers those are placed diagonally.

The **first diagonal** is just only 1s.

The **second diagonal** has counting numbers i.e. 1, 2, 3, 4, 5, 6, 7 etc., which is a sequence of integers.

The **third diagonal** of Pascal’s triangle has sequence of numbers 1, 3, 6, 10, 15, 21 etc. This series of numbers is also called **triangular numbers**.

The **fourth diagonal** is also an interesting sequence of numbers 1, 4, 10, 20, 35 etc., which are called as **tetrahedral numbers**.

## Number pattern on rows of Pascal’s triangle

The sum of numbers on rows of Pascal’s triangle makes up the power of 2.

Let’s understand it from the above figure.

1^{st} row: 1 = 1 = 1^{1}

2^{nd} row: 1 + 1 = 2 = 2^{1}

3^{rd} row: 1 + 2 + 1 = 4 = 2^{2}

4^{th} row: 1 + 3 + 3 + 1 = 8 = 2^{3}

5^{th} row: 1 + 4 + 6 + 4 + 1 = 16 = 2^{4}

6^{th} row: 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^{5}

7^{th} row: 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^{6}

8^{th} row: 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128 = 2^{7}