Question

Show that the binomial coefficients$\left(\begin{array}{l}-1\\ n\end{array}\right)=(-1{)}^n$

Short answer

The statement has been proven.

Step-by-step solution

Given Information  

The binomial series

Definition of the binomial series

The Taylor series for the function given by is the binomial series, where is an arbitrary complex number.

Prove the statement

The binomial series states that $(1+x{)}^p=\sum _{n=0}^{\infty }\left(\begin{array}{l}p\\ n\end{array}\right)x^n$

The formula states that $\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{p(p-1)(p-2)\dots (p-n+1)}{n!}$

$(1+x{)}^{-1}=\sum _{n=0}^{\infty }\left(\begin{array}{c}-1\\ n\end{array}\right)x^n$

$=1-x+\frac{(-1)(-2)}{2!}{x}^2+\frac{(-1)(-2)(-3)}{3!}{x}^3+\dots$

$=1-x+x^2-x^3+\dots$

$=\sum _{n=0}^{\infty }(-1{)}^n{x}^n$

${\text{Hence}}_{\alpha ,}\sum _{n=0}^{\infty }\left(\begin{array}{c}-1\\ n\end{array}\right)x^n=\sum _{n=0}^{\infty }(-1{)}^n{x}^n\text{.}$

The statement has been proven.