Use (1+x)n=1+nx+n(n−1)2!x2+⋯(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots(1+x)n=1+nx+2!n(n−1)x2+⋯ valid for ∣x∣<1|x|<1∣x∣<1.
