Newton Tổng Quát(1+x)α=∑k=0∞(αk)xk,∣x∣<1(1+x)^\alpha = \sum_{k=0}^{\infty}\binom{\alpha}{k}x^k, \quad |x|<1(1+x)α=k=0∑∞(kα)xk,∣x∣<1(αk)=α(α−1)⋯(α−k+1)k!\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}(kα)=k!α(α−1)⋯(α−k+1)Ví dụ: (1+x)1/2=1+x/2−x2/8+...(1+x)^{1/2} = 1 + x/2 - x²/8 + ...(1+x)1/2=1+x/2−x2/8+...👉 Khai triển trên AhaStep