find the first four terms of the binomial expansion of (1-x)^-5

Question

anonymous · Answer

as in power series expansion?

anonymous · Answer

yes please

anonymous · Answer

$$f(x)=\frac{1}{(1-x)^5}$$we need 
$$f(0), f'(0), f''(0), f^{(3)}(0),f^{(4)}(0)$$
\

anonymous · Answer

i don't think there is a snap way to do it.

anonymous · Answer

$$f(0)=1$$
$$f'(x)=5(1-x)^{-6}$$ so 
$$f'(0)=5$$
$$f''(x)=30(1-x)^{-7}$$ so 
$$f''(0)=30$$ 
$$f^{(3)}(x)=210(1-x)^{-8}$$ and 
$$f^{(3)}(0)=210$$

anonymous · Answer

expansion coefficients are 
$$\frac{f^{(n)}(0)}{n!}$$ so we get 
$$1+5x+15x^2+35x^4+$$

anonymous · Answer

oh you have one more but it is clear right? next one will be 
$$\frac{8\times 210}{4!}$$

anonymous · Answer

i get 70

anonymous · Answer

Been struggling with this for awhile now , thanks alot, appreciate it :)

anonymous · Answer

yw hope steps  are clear

anonymous · Answer

yes, thanks, it makes more sense to me now.