Simplify the sum

Question

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 $a_i$ .

Answer 1

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 $2n-2$.