1. ## Calculus: Taylor Series

Find the Taylor series for the function f(x) = 1/((1-x)^2) valid for x near zero, using the fact that 1/(1-x) = the sum from k=0 to infinity of x^k.

2. Differentiate $\frac{1}{1-x}$ and see what you get...

Apply the same to the sum you're given

Differentiate $\frac{1}{1-x}$ and see what you get...
The derivative of $x^n$ is $nx^{n-1}$ so yeah that 1+K one you got is better. (just let n=k+1)