Mathematics identic with numbers. Some of them have special pattern. It is not only arithmetic and geometry sequences but also truly special pattern. One of them is Pascal’s triangle.

In real life it can be applied to solve combination of head and tail problems, polynomials problem (binomial expansion), etc..

Table of Contents

**Definition of Pascal’s Triangle**

Pascal’s triangle was named by a French mathematician, Blaise Pascal (1623-1662). It is one of mathematics concept in algebra. Pascal’s triangle is binomial coefficient (number pattern) in triangle array.

**Pascal’s Triangle pattern**

There are some patterns of Pascal’s triangle.

**1. Diagonal patterns**

There are 6 rows of Pascal’s triangle. Look at the diagonals of Pascal’s triangle.

- First diagonal is always “1” (the blue ones)
- Second diagonal is counting numbers “1,2,3,…” (red ones)
- Third diagonal is triangular numbers “1,3,6,…”(purple ones)

**2. Fibonacci sequences**

Fibonacci sequences is 0, 1, 1, 2, 3, 5, 8, … Pascal’s triangle has it in addition some of the numbers.

Starting with 0, then Pascal’s triangle has the remainder.

Look at each color. Add the numbers that have same color.

1

1

1 + 1 = 1

1 + 2 = 3

1 + 3 + 1 = 5

1 + 4 + 3 = 8

…

It will be Fibonacci sequences.

**3. Squares**

Look at the second diagonal of Pascal’s triangle.

The square of the numbers is equals to addition the numbers next to it and below.

Some of the squares are

2^{2} = 4 = 3 + 1

4^{2} = 16 = 10 + 6

**4. Exponents of 11**

Look at each row. It is the result of exponents (power) of 11.

11^{0} = 1

11^{1} = 11

11^{2} = 121

11^{3} = 1331

11^{4} = 14641

But 11^{5} ≠ 15101051.

11^{5} = 161051. It is because 15101051 has more digit numbers. Then it must be 1(5+1)(0+1)051 = 161051.

It also has same concept to 11^{6} and bigger power.

**5. Horizontal sum**

Pay attention in each row.

Each row is the result of 2^{n}.

2^{0} = 1

2^{1} = 2 = 1+ 1

2^{2} = 4 = 1 + 2 + 1

2^{3} = 8 = 1 + 3 + 3 + 1

2^{4} = 16 = 1 + 4 + 6 + 4 + 1

2^{5} = 32 = 1 + 5 + 10 + 10 + 5 + 1

**6. Symmetrical numbers**

Another pattern of Pascal’s triangle is symmetrical number between left and right side. Look at the triangle and see how the mirror of the numbers.

**Pascal’s Triangle formula**

Pascal’s triangle has many numbers. If there are five rows you can determine the numbers in 8^{th} rows or others. It uses the formula (combination concept).

Note:

- n : row
- r : term / element, 0 ≤ r ≤ n

if there are only six rows then the problem is determining the number in 10^{th} row term 4,

Triangle array in Pascal’s triangle is arranged by summing adjacent elements in previous row. It starts with “1” in zero row then continued with 1 and 1 in first row. Second row is addition of number in first row and continued with same step in next row.

Why there is 0^{th}?

The reason is making Pascal’s triangle (elements in the array) correspond to binomial coefficients.

Pascal’s triangle relates to polynomial. It is as coefficient of binomial expansion.

- (x+1)
^{2}= 1x^{2}+ 2x + 1x^{0}= x^{2}+2x + 1 - (x+1)
^{3}= 1x^{3}+ 3x^{2}+ 3x + 1x^{0}= x^{3}+ 3x^{2}+ 3x + 1 - (x+1)
^{4}= 1x^{4}+ 4x^{3}+ 6x^{2}+ 4x + 1x^{0}= x^{4}+ 4x^{3}+ 6x^{2}+ 4x + 1 - Etc.

Look at the power and the pascal’s triangle. The power shows the number of rows of pascal’s triangle.

**Examples**

1. Determine the value of

2. What is row 15, term 3 in Pascal’s triangle?

Row 15 term 3 means

3. What is the binomial expansion of (x+1)^{6}?

Using pascal’s triangle in 6^{th} row that is 1 6 15 20 15 6 1 then

(x+1)^{6} = 1x^{6} + 6x^{5} + 15x^{4}+ 20x^{3} + 15x^{2} + 6x + 1

= x^{6} + 6x^{5} + 15x^{4}+ 20x^{3} + 15x^{2} + 6x + 1

4. What is the binomial expansion of (m+2)^{5}?

Using pascal’s triangle in 5^{th} row.

1 5 10 10 5 1

and

(a+b)^{5} = 1a^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} + 5ab^{4} + 1b^{5}

Then

(m+2)^{5} = 1m^{5} + 5m^{4}(2) + 10m^{3}(2)^{2} + 10m^{2}(2)^{3} + 5m(2)^{4} + 1(2)^{5}

= 1m^{5} + 10m^{4} + 40m^{3} + 80m^{2} + 80m + 32

5. What is the binomial expansion of (a-2)^{7}?

Using 7^{th} row

1 7 21 35 35 21 7 1

then

(a-2)^{7} = 1a^{7} + 7a^{6}.(-2) + 21a^{5}.(-2)^{2} + 35a^{4}.(-2)^{3} + 35a^{3}.(-2)^{4} + 21a^{2}.(-2)^{5} + 7a.(-2)^{6} + 1.(-2)7

= a^{7} – 14a^{6} + 84a^{5} – 280a^{4} + 140a^{3} – 672a^{2} + 448a – 14