Determine the polynomial $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$$ by using the binomial theorem.
My issue with this question is taking the $n$th derivative after using the binomial theorem. I understand what will happen but I don't know how to set it out properly with the sigma so that is it clear to the reader. Can anyone provide a model answer?
By the binomial theorem and the linearity of the differential operator, we have that $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}\left(\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}x^{2k}\right)=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{n-k}}{n!}\frac{d^n\left(x^{2k}\right)}{dx^n}.$$ Can you take it from here?