Show that:
$$P_{2n}^1(0)=0$$
$$P_{2n}^1(0)=(-1)^n\frac{(2n+1)!}{(2^nn!)^2} $$
by each of three methods:
a) use of recurrence relations $$(2n+1)(1-x^2)^{\frac{1}{2}}P_{n}^m=(n+m)(n+m-1)P_{n-1}^{m-1}-(n-m+1)(n-m-2)P_{n+1}^{m-1}$$ b) expansion of generating function $$\frac{(2m)!(1-x^2)^{m/2}}{2^mm!(1-2tx+t^2)^{m+1/2}}=\sum_{s=0}^\infty P_{s+m}^m(x)t^s$$ c) Rodrigues formula $$P_n(x)=\frac{1}{n!\space2^n}\frac{d^n}{dx^n}\{(x^2-1)^n\} $$
