Simplify the sum Consider the sum of two polynomials
$$
\sum_{k=0}^{n-1} {{{n-1} \choose {k}}^2 z^{2k}}+\sum_{k=0}^{n-2} {{n-2} \choose {k}} {{n} \choose {k+1}} z^{2k+1}=\sum_{i=0}^{2n-1}a_i z^i.
$$
I want to find the exact expression for $a_i$. Of course I may divide  it into even and odd parts but I hope  there is a formula for $a_i$ .


A sum $A(z)=\sum_{k=0}^{2n}a_kz^k$ can be split into even and odd parts via
  \begin{align*}
\frac{1}{2}\left(A(z)+A(-z)\right)
&=\frac{1}{2}\left(\sum_{k=0}^{2n}a_kz^k+\sum_{k=0}^{2n}a_k(-z)^k\right)\\
&=\sum_{k=0}^{2n}\frac{1+(-1)^k}{2}a_kz^k\\
&=\sum_{k=0}^na_{2k}z^{2k}
\end{align*}
  and
  \begin{align*}
\frac{1}{2}\left(A(z)-A(-z)\right)=\sum_{k=0}^{2n}\frac{1-(-1)^k}{2}a_kz^k
=\sum_{k=0}^{n-1}a_{2k+1}z^{2k+1}
\end{align*}

Therefore the LHS of your expression can be written as
\begin{align*}
\sum_{k=0}^{n-1}\binom{n-1}{k}^2&z^{2k}+\sum_{k=0}^{n-2}\binom{n-2}{k}\binom{n}{k+1}z^{2k+1}\\
&=\sum_{k=0}^{2n-2}\left(\binom{n-1}{k}^2\frac{1+(-1)^k}{2}
+\binom{n-2}{k}\binom{n}{k+1}\frac{1-(-1)^k}{2}\right)z^k
\end{align*}
and could now be transformed in order to find a convenient representation.
Note: On your RHS you could replace the upper bound of the sum with $2n-2$.
