How find this sum $\sum\limits_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$ 
Find the sum close form
  $$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$

if we let
$$\dfrac{x}{2}=y$$
then 
$$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$
this PDF  have this page 5
$$\sum_{k=j}^{n}\binom{n}{k}\binom{k}{j}=2^{n-j}\binom{n}{j}$$
the solution can see
page 5
Maybe my problem can use this mathods?Thank you

Hint: A classical method is to specialize the general identity $$(1+a+b)^{2n}=\sum_{i,j}{2n\choose i,j}a^ib^j$$ to $a=ty$ and $b=y/t$. One gets $$(1+ty+y/t)^{2n}=\sum_{i,j}{2n\choose i,j}t^{i-j}y^{i+j},$$ and in particular $$[t^0](1+ty+y/t)^{2n}=\sum_{i}{2n\choose i,i}y^{2i}=\sum_{i}{2n\choose 2i}{2i\choose i}y^{2i}=f(y),$$ or, equivalently, $$f(y)=y^{2n}\cdot[t^{2n}](t^2+t/y+1)^{2n}.$$
Can you continue?
