I have to prove two statements by induction but I really don't know how! Could somebody give me some advice?
1) $\left(\begin{array}{c} -\frac{1}{2} \\ n \end{array}\right) = \Large{(-1)^n \frac{(2n-1)!!}{(2n)!!}} \forall n \in \mathbb{N} $
2) $\left(\begin{array}{c} \frac{1}{2} \\ n \end{array}\right) = \Large{(-1)^{n+1} \frac{(2n-3)!!}{(2n)!!}} \forall n \in \mathbb{N}-{1} $
The definition of the binomial coefficient is: $$ {\alpha\choose n}=\frac{\alpha(\alpha-1)...(\alpha-n+1)}{n!} $$ From this definition, it is clear that $$ {\alpha\choose n+1}=\frac{\alpha-n}{n+1}{\alpha\choose n} $$ For $\alpha=-1/2$, this gives $$ {-1/2\choose n+1}=\frac{-1/2-n}{n+1}{-1/2\choose n}=-\frac{2n+1}{2n+2}{-1/2\choose n} $$ By the induction hypothesis: $$ -\frac{2n+1}{2n+2}{-1/2\choose n}=-\frac{2n+1}{2n+2}(-1)^n\frac{(2n-1)!!}{(2n)!!}=(-1)^{n+1}\frac{(2n+1)!!}{(2n+2)!!} $$ This completes the induction step and thus proves the formula $\forall n\in\mathbb{N}$.
The second exercise proceeds completely analogously.