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
 A: Given any formal Laurent series $\;(???) = \sum \alpha_{k_1 k_2 \ldots k_n} t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n}$, we will use the notation $[ t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n} ](???)$ to denote the coefficient $\alpha_{k_1 k_2 \cdots k_n}$ in front of corresponding monomial. 
Instead of $f(y)$, let us denote the polynomial we wish to find a closed form as $p_{2n}(y)$. We have
$$\begin{align}
p_{2n}(y) 
&= \sum_{i=0}^{n}\binom{2n}{2i}\binom{2i}{i} y^{2i}
= \sum_{i=0}^{n} \binom{2n}{2i} y^{2i}\bigg( [t^0](t + t^{-1})^{2i}\bigg)\\
&= \sum_{i=0}^{2n} \binom{2n}{i} \bigg( [t^0](y(t + t^{-1}))^i\bigg)
= [\;t^0\;] \bigg( 1 + y(t+t^{-1})\bigg)^{2n}
\end{align}
$$
Substitute $t$ by $e^{i\theta}$ in above formal expression and notice for any $k \in \mathbb{Z}$, we have
$$\frac{1}{2\pi}\int_0^{2\pi} e^{ik\theta} d\theta = \begin{cases}1,&k = 0\\0,&\text{ otherwise }\end{cases}$$
We obtain an integral representation for $p_{2n}(y)$,
$$p_m(y) = \frac{1}{2\pi}\int_0^{2\pi} (1+2y\cos\theta)^{m} d\theta\quad\text{ for }\quad m = 2n$$
Treat this as a definition for $p_m(y)$ for general $m \in \mathbb{N}$ and consider following generating function:
$$p(y,\rho) = \sum_{m=0}^\infty p_m(y)\rho^m$$
It is easy to see
$$
p(y,\rho)
= \frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-\rho(1+2y\cos\theta)}
= \frac{1}{4\pi y\rho}\int_0^{2\pi}\frac{d\theta}{\frac{1-\rho}{2y\rho}-\cos\theta}
$$
Using the identity
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{a - \cos\theta} = \frac{1}{\sqrt{a^2-1}}\quad\text{ for } a > 1$$
We get
$$\begin{align}
p(y,\rho) &= \frac{1}{2y\rho}\frac{1}{\sqrt{\left(\frac{1-\rho}{2y\rho}\right)^2 - 1}}
= \frac{1}{\sqrt{1-2\rho + (1- 4y^2)\rho^2}}\\
&= \frac{1}{\sqrt{1-2\frac{1}{\sqrt{1-4y^2}}(\rho\sqrt{1-4y^2}) + (\rho\sqrt{1-4y^2})^2}}\
\end{align}
$$
Compare this with the generating function for Legendre polynomials,
$$\frac{1}{\sqrt{1-2zt+t^2}} = \sum_{k=0}^\infty P_k(z) t^k$$
We find
$$p(y,\rho) = \sum_{k=0}^\infty P_k\left(\frac{1}{\sqrt{1-4y^2}}\right) \left(\rho\sqrt{1-4y^2)}\right)^k$$
This leads to the expression we claimed in comment:
$$p_{2n}(y) = (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)$$
For example, when $y = \frac12$, this leads to an interesting identity:
$$\begin{align}
\sum_{i=0}^{n} \frac{\binom{2n}{2i}\binom{2i}{i}}{2^{2i}} 
&= p_{2n}\left(\frac12\right)
= \lim_{y\to\frac12^{-}} (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)\\
&= [ t^{2n} ] P_{2n}(t) = \frac{(4n-1)!!}{(2n)!}
\end{align}
$$
A: 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?
