I had a question: To find the statistic's average of the UDs (Up-Down) in Motzkin paths. I solved it but finally i got something in the statistic that I tried a lot to simplify but did not succeed unfortunately! I got that the statistic = $$\dfrac{\sum_{i=0}^{n/2} c_i(2i+1) {\binom{(n+1)}{(n-2i)}}}{\sum_{i=0}^{n/2} c_i {n \choose (n-2i)}}$$ $c_i$ is the $i$-th Catalan number. How can I simplify that?
How can I get the variance by using this complicated formula?
Given OPs expression we can find the sequence of numbers of the numerator \begin{align*} &\left(\sum_{j=0}^{\lfloor n/2\rfloor}\frac{2j+1}{j+1}\binom{2j}{j}\binom{n+1}{n-2j}\right)_{n\geq 0}\\ &\qquad=\left(1,2,6,16,45,126,357,1\ 016,2\,907,8\,350,24\,068,\ldots\right) \end{align*} archived as A005717 in OEIS as well as the sequence of numbers of the denominator \begin{align*} &\left(\sum_{j=0}^{\lfloor n/2\rfloor}\frac{1}{j+1}\binom{2j}{j}\binom{n}{n-2j}\right)_{n\geq 0}\\ &\qquad=\left(1,1,2,4,9,21,51,127,323,835,2\,188,5\,798,\ldots\right) \end{align*} archived as A001006 in OEIS.
Since both entries do not provide a closed formula we rather don't expect a simplification.