Series with Binomial probability

Question

I found the following formula in the solution of an exercise and I'm not able to understand where it come from and if it is correct. $$\sum_{x=0}^\infty\binom{n}{x} p^x(1-p)^{-x} = \left(1+\frac{p}{1-p}\right)^n $$

Could someone explain me how to get such a result (if is correct)?

Answer 1

It comes from the Binomial theorem. If you have $(1+a)^n=\sum _{k=0}^n\binom{n}{k}a^k.$ Take $a=p(1-p)^{-1}.$ Notice also that $\binom{n}{k}=0$ if $k>n$(How can you select more objects that you originally have?).

Answer 2

It is correct. $$\sum_{x=0}^\infty\binom{n}{x} p^x(1-p)^{-x}=(1-p)^{-n}\sum_{x=0}^n\binom{n}{x} p^x(1-p)^{n-x}=(1-p)^{-n}(p+(1-p))^n=(1-p)^{-n}$$where $\binom{n}x:=0$ if $x>n$.

Further: $$1+\frac{p}{1-p}=(1-p)^{-1}$$