Let $P_n(x) = \dfrac{1}{2^n n!}\dfrac{d^n}{dx^n} [(x^2-1)^n]$, we know that $\sum\limits_{n=0}^{\infty} P_n(x)t^n=(1-2tx+t^2)^{-1/2}$
How can we proved
$P_{2n}(0)=\dfrac{(-1)^n(2n)!}{4^n(n!)^2}$ and $P_{2n+1}(0)=0$
I tried to set $x=0$, then we get $f(t)=\sum\limits_{n=0}^{\infty} P_n(0)t^n=(1+t^2)^{-1/2}$, we then have $f^{(n)}(0)=n! P_n(0)$. But I do not know how to get the required answer by differentiating $(1+t^2)^{-1/2}$ $n$ times. Any ideas? (any answers using Rodrigues's formula would also be appreciated, though I do not see any obvious way of doing that)
By the binomial theorem: $$ f(t) = (1 + t^2)^{-1/2} = \sum_{n \ge 0} \binom{-1/2}{n} t^{2 n} $$ The respective binomial coefficient is: $$ \binom{-1/2}{n} = \frac{(-1)^n}{2^{2n}} \binom{2n}{n} $$ so you have your result.