Pascal's Triangle

Pascal's Triangle is a triangular arrangement of numbers where each number is the sum of the two numbers directly above it in the previous row. It is named after the French mathematician Blaise Pascal, although it was studied by mathematicians in various cultures long before him.

Structure of Pascal's Triangle

• The triangle starts with a 1 at the top.
• Each row begins and ends with 1.
• Any other number in the triangle is the sum of the two numbers immediately above it in the previous row.

History of Pascal's Triangle

Pascal's Tringale is named after the French mathematician Blaise Pascal, though it was known to mathematicians in ancient India, China, and Persia long before Pascal's time. The earliest known version appeared in China around 1000 BCE, in the works of the mathematician Jia Xian. Later, the Persian mathematician Al-Karaji contributed to its development in the 10th century. The triangle became widely recognized in the West due to Pascal's work in the 17th century, though he did not invent it.

Pascal’s Triangle Patterns

We observe various patterns in Pascal’s triangle they are:

• Diagonals in Pascal’s Triangle
• Binomial Coefficients
• Horizontal Sum
• Prime Numbers in Triangle
• Fibonacci Pattern, etc.

Diagonals in Pascal’s Triangle

Each rightward diagonal of Pascal’s Triangle, when considered as a sequence represents the different numbers such as the first rightward diagonal represents a sequence of number 1, the second rightward diagonal represents triangular numbers, the third rightward diagonal represents the tetrahedral numbers, the fourth rightward diagonal represents the Penelope numbers and so on.

Pascal’s Triangle Binomial Expansion

We can easily find the coefficient of the binomial expansion using Pascal's Triangle. The elements in the (n+1)th row of the Pascal triangle represent the coefficient of the expanded expression of the polynomial (x + y)ⁿ.

We know that the expansion of (x + y)ⁿ is,
(x + y)ⁿ = a₀xⁿ + a₁x^n-1y + a₂x^n-2y² + … + a_n-1xy^n-1 + a_nyⁿ

Here, a₀, a₁, a₂, a₃, ...., a_n are the term in the (n+1)th row of Pascal's Triangle

Horizontal Sum of Rows

On close observing Pascal’s Triangle we can conclude that the sum of any row in Pascal’s triangle is equal to a power of 2. The formula for the same is, For any (n + 1)^th row in Pascal’s Triangle the sum of all the elements is, 2ⁿ

Applying this Formula in the first 4 rows of Pascal’s triangle we get,

1 = 1 = 2⁰
1 + 1 = 2 = 2¹
1 + 2 + 1 = 4 = 2²
1 + 3 + 3 + 1 = 8 = 2³

Exponents of 11

Each row of Pascal's Triangle corresponds to a power of 11:

Row n (starting from n = 0) represents 11ⁿ.

Examples:

1. Row 0: 1 → 11⁰ = 1
2. Row 1: 1, 1 → 11¹ = 11
3. Row 2: 1, 2, 1 → 11² = 121
4. Row 3: 1, 3, 3, 1 → 11³ = 1331

Note: For higher powers of 11, the digits in the triangle rows exceed single digits, causing a carry-over effect. In such cases, the numbers in Pascal's Triangle no longer directly represent 11ⁿ. Instead, the pattern can still be reconstructed by summing the digits appropriately.

Prime Numbers in Pascal’s Triangle

Another very interesting pattern in the Pascals triangle is that if a row starts with a prime number (neglecting 1 at the start of each row), then all the elements in that row are divisible by that prime number. This pattern does not hold true for the composite numbers. 

For example, the eighth row in the Pascal triangle is,

1    7    21    35    35    21    7    1

Here, all the elements are divisible by 7.

Fibonacci Sequence in Pascal’s Triangle

We can easily obtain the Fibonacci sequence by simply adding the numbers in the diagonals of Pascal's triangle. This pattern is shown in the image added below,

Read More: Interesting Facts about Pascal's Triangle

Pascal’s Triangle Formula

Pascal Triangle Formula is the formula that is used to find the number to be filled in the m^th column and the n^th row. As we know the terms in Pascal's triangle are the summation of the terms in the above row. So we require the elements in the (n-1)th row, and (m-1)th and nth columns to get the required number in the mth column and the nth row.

The elements of the nth row of Pascal's triangle are given, ⁿC₀, ⁿC₁, ⁿC₂, ..., ⁿC_n. 

The formula for finding any number in Pascal's triangle is:

ⁿCm = ^n-1C_m-1 + ^n-1C_m

Where,

• ⁿC_m represents the (m+1)th element in the nth row., and
• n is a non-negative integer [0 ≤ m ≤ n]

How to Use Pascal’s Triangle?

We use the Pascal triangle to find the various cases of the possible outcomes in probability conditions. This can be understood by the following example, tossing a coin one time we get two outcomes i.e. H and T this is represented by the element in the first row of Pascal's Triangle.

Similarly tossing a coin two times we get three outcomes i.e. {H, H}, {H, T}, {T, H}, and {T, T} this condition is represented by the element in the second row of Pascal's Triangle.

Thus, we can easily tell the possible number of outcomes in tossing a coin experiment by simply observing the respective elements in the Pascal Triangle.

The table below tells us about the cases if a coin is tossed one time, two times, three times, and four times, and its accordance with Pascal's Triangle

Pascal’s Triangle Properties

Various Properties of Pascal's Triangle are,

• Every number in the Pascal triangle is the sum of the number above it.
• The starting and the end numbers in Pascal's triangle are always 1.
• The first diagonal in Pascal's Triangle represents the natural number or counting numbers.
• The sum of elements in each row of Pascal's triangle is given using a power of 2.
• Elements in each row are the digits of the power of 11.
• The Pascal triangle is a symmetric triangle.
• The elements in any row of Pascal's triangle can be used to represent the coefficients of Binomial Expansion.
• Along the diagonal of Pascal's Triangle, we observe the Fibonacci numbers.

Articles related to Pascal's Triangle:

• Binomial Theorem
• Binomial Random Variables and Binomial Distribution

For Programmers: Program to print to Pascal's Triangle

Solved Examples of Pascal’s Triangle

Example 1: Find the fifth row of Pascal’s triangle. 

Solution:

The Pascal triangle with 5 row is shown in the image below,

Example 2: Expand using Pascal Triangle (a + b)². 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)² = c₀a²b⁰ + c₁a¹b¹ + c₂a⁰b² 

Now let’s build a Pascal’s triangle for 3 rows to find out the coefficients.  

The values of the last row give us the value of coefficients, c₀ = 1, c₁ = 2, c₂ =1

 (a + b)² = a²b⁰ + 2a¹b¹ + a⁰b² 

Thus verified. 

Example 3: Expand using Pascal Triangle (a + b)⁶. 

Solution:

First write the generic expressions without the coefficients.  

 (a + b)⁶ = c₀a⁶b⁰ + c₁a⁵b¹ + c₂a⁴b² + c₃a³b³ + c₄a²b⁴ + c₅a¹b⁵ + c₆a⁰b⁶  

Now let’s build a Pascal’s triangle for 7 rows to find out the coefficients.  

The values of the last row give us the value of coefficients. 

c₀ = 1, c₁ = 6, c₂ = 15, c₃ = 20, c₄ =15, c₅ = 6 and c₆ = 1. 

(a + b)⁶ = 1a⁶b⁰ + 6a⁵b¹ + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6a¹b⁵ + 1a⁰b⁶  

Example 4: Find the second element in the third row of Pascal’s triangle.

Solution:

To find the 2nd element in the 3rd row of Pascal’s triangle.

We know that the n^th row of Pascal’s triangle is ⁿC₀, ⁿC₁, ⁿC₂, ⁿC_3…

The Pascal Triangle Formula is, ⁿC_k = ^n-1C_k-1  + ^n-1C_k,where ⁿC_k represent (k+1)^th element in n^th row.

Thus, 2nd element in the 3rd row is,

³C₁ = ²C₀ + ²C₁
= 1 + 2
= 3

Thus, the second element in the third row of Pascal’s triangle is 3.