Generate pascal's triangle up to n=5 and use it to expand (2x-3y)⁵?

To generate Pascal's triangle up to n=5, we start with the first row, which is always 1. Then, for each subsequent row, we add the two numbers above it to get the new number. Here is the Pascal's triangle up to n=5:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

Now, let's expand (2x-3y)⁵ using Pascal's triangle:

(2x-3y)⁵ = 1(2x)⁵(-3y)⁰ + 5(2x)⁴(-3y)¹ + 10(2x)³(-3y)² + 10(2x)²(-3y)³ + 5(2x)¹(-3y)⁴ + 1(2x)⁰(-3y)⁵

Simplifying each term:

(2x)⁵ = 32x⁵
(-3y)⁰ = -3⁰ = 1
(2x)⁴ = 16x⁴
(-3y)¹ = -3y
(2x)³ = 8x³
(-3y)² = 9y²
(2x)² = 4x²
(-3y)³ = -27y³
(2x)¹ = 2x
(-3y)⁴ = 81y⁴
(2x)⁰ = 1
(-3y)⁵ = -243y⁵

Now, substituting these values into the expanded form:

(2x-3y)⁵ = 1(32x⁵)(1) + 5(16x⁴)(-3y) + 10(8x³)(9y²) + 10(4x²)(-27y³) + 5(2x)(81y⁴) + 1(1)(-243y⁵)

Simplifying further:

(2x-3y)⁵ = 32x⁵ - 240x⁴y + 720x³y² - 1080x²y³ + 810xy⁴ - 243y⁵

Therefore, the expansion of (2x-3y)⁵ is 32x⁵ - 240x⁴y + 720x³y² - 1080x²y³ + 810xy⁴ - 243y⁵.