Question

The last four terms of the expansion of the binomial (2x³ + 3y²)⁷ are ________.

1) 128x²¹ - 1344x¹8y² + 6048x¹⁵y⁴ - 15120x²y⁶
2) 128x²¹ + 1344x¹8y² - 6048x¹⁵y⁴ + 15120x²y⁶
3) 128x²¹ + 1344x¹8y² + 6048x¹⁵y⁴ + 15120x²y⁶
4) 128x²¹ - 1344x¹8y² - 6048x¹⁵y⁴ - 15120x²y⁶

Answer 1

The last four terms of the expansion of the binomial (2x³ + 3y²)⁷ are 128x²¹ + 1344x¹8y² - 6048x¹⁵y⁴ + 15120x²y⁶.

Step-by-step explanation:

The last four terms of the expansion of the binomial (2x³ + 3y²)⁷ are 128x²¹ + 1344x¹8y² - 6048x¹⁵y⁴ + 15120x²y⁶.

The binomial theorem states that the expansion of (a + b)ⁿ can be found using the formula: (a + b)ⁿ = C(n, 0) * aⁿ * b⁰ + C(n, 1) * aⁿ⁻¹ * b¹ + ..., where C(n, k) represents the binomial coefficient and is equal to n! / (k! * (n - k)!).

Using this formula, we can expand (2x³ + 3y²)⁷ and find the last four terms as follows:

1. Term 1: C(7, 3) * (2x³)⁴ * (3y²)³ = 35 * 16x¹² * 27y⁶ = 15120x¹²y⁶
2. Term 2: C(7, 2) * (2x³)⁵ * (3y²)² = 21 * 32x¹⁵ * 9y⁴ = 6048x¹⁵y⁴
3. Term 3: C(7, 1) * (2x³)⁶ * (3y²)¹ = 7 * 64x¹⁸ * 3y² = 1344x¹⁸y²
4. Term 4: C(7, 0) * (2x³)⁷ * (3y²)⁰ = 1 * 128x²¹ * 1 = 128x²¹