Find the first three terms of the expansion of (1 + 2x)−3 and state when it is valid.
- Use n = −3 and replace x by 2x: 1 + (−3)(2x) + [(−3)(−4)/2!](2x)2.
- Second term: (−3)(2x) = −6x.
- Third term: (12 ÷ 2) × 4x2 = 6 × 4x2 = 24x2.
- Validity: |2x| < 1, so |x| < 1/2.
Answer: 1 − 6x + 24x2, valid for |x| < 1/2.