For any n, (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + … When n is not a positive whole number, the series goes on forever.
It is valid only when |x| < 1. For (1 − 4x)^(−1/2), the 'x' is −4x, so it is valid when |4x| < 1, which means −1/4 < x < 1/4. For (2 + x)ⁿ, take out 2ⁿ first to get the form (1 + x/2)ⁿ.
Worked example
Find the first three terms of (1 + x)⁻¹.
- n = −1: 1 + (−1)x + (−1)(−2)x²/2.
- = 1 − x + x².
Answer:1 − x + x²
Key words
- validity
- the x values where the series works
- factorial (!)
- n! = n × (n − 1) × … × 1
- infinite series
- a sum that never ends
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