Is it possible to get a closed form for coefficients of $$\left(1+\frac{2t}{(1-t)^2}\right)^{-n}$$ there $n$ - positive integer?
It's easy to obtain the formula for $m$-th coefficient as $$\binom{n}{1}\binom{m}{m-1}(-2)+\binom{n+1}{2}\binom{m+1}{m-2}(-2)^2+\dots$$ but I can't simplify it further.
An idea to make, possibly, things easier:
You can make some algebraic simplifications. Choose to develop whatever you like better:
$$\left(1+\frac{2t}{(1-t)^2}\right)^{-n}=\left(\frac{(1-t)^2}{1+t^2}\right)^n=\left(1-\frac{2t}{1+t^2}\right)^n=\sum_{k=0}^n\binom nk(-1)^k\left(\frac{2t}{1+t^2}\right)^k$$