Explanation for binomial sums $\sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1}$

Question

I was looking at some negative binomial coefficient problems and I stumbled upon this explanation

$$\sum_{n=0}^{\infty} \binom{n+2}{3} x^n = \sum_{n=0}^{\infty} \binom{n+2}{n-1} x^n= \sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1} $$

I was wondering how the author arrived at the conclusion that

$$\sum_{n=0}^{\infty} \binom{n+2}{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n-1} (-1)^{n-1} x^n = \sum_{n=0}^{\infty} \binom{-4}{n} (-1) x^{n+1}$$

Answer 1

Index transformation (replace $n$ by $n+1$) gives $$ \sum_{n=0}^\infty \binom{-4}{n-1}(-1)^{n-1} x^n = \sum_{n=-1}^\infty \binom{-4}{n}(-1)^n x^{n+1} $$ Now, on the right hand side, the term for $n=-1$, namely $$ \binom{-4}{-1} (-1)^{-1} x^0 = 0 $$ is zero due to $\binom{-4}{-1} = 0$. Hence $$ \sum_{n=0}^\infty \binom{-4}{n-1}(-1)^{n-1} x^n = \sum_{n=-1}^\infty \binom{-4}{n}(-1)^n x^{n+1} = \sum_{n=0}^\infty \binom{-4}{n}(-1)^n x^{n+1} $$