## Find Taylor Polynomial order 5 f$f(x) = \frac{1}{(1-x)}$, a = 0

Find Taylor Polynomial order 5 $$f(x) = \frac{1}{(1-x)}$$ , a = 0

So i start of with: f(x) = $$f(0) = \frac{1}{(1-0)}$$ –> 1

f'(x) = $$f(0) = \frac{1}{(1-0)^2 }$$ –> 1

f”(x) = $$f(0) = \frac{2}{(1-0)^3 }$$ –> 2

f”'(x) = $$f(0) = \frac{6}{(1-0)^4 }$$ –> 6

f””(x) = $$f(0) = \frac{24}{(1-0)^5 }$$ –> 24

f””(x) = $$f(0) = \frac{120}{(1-0)^6 }$$ –> 120

$$1 + x + \frac{2x^2}{2} + \frac{6x^3}{3} + \frac{24x^4}{4} + \frac{120x^5}{5}$$

Answer = $$1 + x + x^2 + 2x^3 + 5x^4 + 24x^5$$

The problem is that a calculator i used to check it up said the answer was

answer = $$1 + x + x^2 + x^3 + x^4 + x^5$$

So im confused, im I right or is the calculator wrong? Thanks!