Pascal's Triangle

Here is an exploration with another type of pattern. This pattern is based on Pascal’s Triangle. As you work through the examples, refer to this diagram.

Notice the patterns in the triangle.

In each row the exponent of a decreases by 1 as you move from left to right. The exponent of b increases by 1 as you move from left to right.

For each term in a row, the sum of the exponents is the same. The sum is one less than the row number.

Also notice the patterns in the coefficients of the terms.

• The terms on the far left and far right always have coefficient 1.

• To find the coefficient of a term inside the triangle, add the coefficients of the two adjacent terms in the row directly above it.

Note:

 

Example 1

Use Pascal’s Triangle to find (a + b)⁵.

Solution

Using the patterns, we can write the expansion of (a + b)⁵.

First, we fill in the exponents:

1a⁵b⁰ + __a⁴b¹ + __a³b² + __a²b³ + __a¹b⁴ + 1a⁰b⁵

Next, we use the coefficients for the expansion of (a + b)⁴ to find the coefficients of (a + b)⁵.

So, the expansion of (a + b)⁵ is: a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

 

Example 2

Use Pascal’s Triangle to find (3x + 2y)⁴.

Solution

This corresponds to the fourth row of the triangle.

Fourth row of the triangle:

(a + b)⁴ = 1a⁴ + 4a³b + 6a²b² + 4ab³ + 1b⁴

Substitute 3x for a and 2y for b:

(3x + 2y)⁴ = 1(3x)⁴ + 4(3x)³(2y) + 6(3x)2(2y)2 + 4(3x)(2y)3 + 1(2y)4

Simplify. = 81x⁴ + 216x³y + 216x²y² + 96xy³ + 16y⁴

Thus, (3x + 2y)⁴ = 81x + 216x³y + 216x²y² + 96xy³ + 16y⁴.