Negative binomial coefficients

By Induction : a) $\binom{-1}{1} = \frac{-1}{1!} = - 1 = (-1)^1 \binom{2\cdot1 - 1}{1}$ $\binom{-2}{2} = \frac{(-2)\cdot(-3)}{2!} = 3 = (-1)^2 \binom{2\cdot2 - 1}{2}$ b) Let's suppose that for $n > 2$ $\binom{-n}{n} = (-1)^n \binom{2n - 1}{n}$ , then $\boxed{1.-}$ $\binom{-n - 1}{n + 1} = \frac{(-n - 1)\cdot (-n - 2) \cdot\cdot \cdot (-n - n - 1)}{(n+1)!} = (-1)^{n + 1} \frac{(n +1) \cdot (n+ 2)\cdot\cdot \cdot (2n + 1)}{(n+1)!} =$ $= (-1)^{n + 1} \binom{2(n + 1) - 1}{n+ 1} = (-1)^{n + 1} \binom{2n + 1}{n + 1}$ . Therefore, $\forall n \in \mathbb{Z}^+ = \{1,2,3,...\}$ , $\binom{-n}{n} = (-1)^n \binom{2n - 1}{n}$

$\boxed{2.-}$ Where have I used induction in step b)?

$\frac{(-n - 1)\cdot (-n - 2) \cdot\cdot \cdot (-n - n - 1)}{(n+1)!} = \frac{(- n)\cdot (-n - 1)\cdot (-n - 2) \cdot\cdot \cdot (-n - n +1)\cdot (-n -n ) \cdot (-n - n - 1)}{(-n)\cdot(n+1)!} =$ Here is the step of induction $= (-1)^{n-1} \binom{2n -1}{n} \cdot \frac{(-2n)(-n -n -1)}{n(n+1)} = (-1)^{n-1} \binom{2n -1}{n} \cdot \frac{(-2)(-n -n -1)}{(n+1)} =$ $= (-1)^{n +1} \binom{2n -1}{n} \cdot \frac{2(2n +1)}{(n+1)} =(-1)^{n +1} \binom{2n +1}{n+1}$ For ending $\large A = \frac{\binom{2n + 1}{n + 1}}{\binom{2n - 1}{n}} = \frac{\frac{(2n + 1)!}{(n+1)! \cdot n!}}{\frac{(2n - 1)!}{n! \cdot (n - 1)!}} = \frac{\frac{(2n + 1)!}{(n+1)!}}{\frac{(2n - 1)!}{(n - 1)!}} = \frac{(2n + 1) \cdot 2n}{(n + 1) \cdot n} = \frac{2(2n +1)}{n + 1}$ $\forall n \in \mathbb{Z}^+ = \{1,2,3,...\}$