R-visible up steps. Similarly, a partial Motzkin path has no L-visible down steps.

We have the following summation formula for weighted free Motzkin paths.

The sum of weights of free $(k-t-1, t)$-Motkzin paths of length $i$ equals $k^i$.

The proof of the above lemma is obvious because of the relation $$(1+k-t-1+t)^i=k^i.$$ We are now led to establish a bijection for the identity (\[mi\]).

\[the.1\] There is a bijection between the set of partial $(k-t-1,t)$-Motzkin paths of length $i$ with an elevation line and the set of free $(k-t-1,t)$-Motzkin path of length $i$.

The bijection is just the elevation operation. The reverse map is also easy. For a free Motzkin path, one can identify the L-visible down steps, if any, then change these L-visible down steps to up steps by elevating their end points.

For a partial $(k-t-1, t)$-Motzkin path $P$ with an elevation line $y=p$, suppose that $Q$ is the elevation of $P$ with respect to $y=p$. It is clear that the weight of $Q$ equals $t^p |P|$. If $P$ ends at level $j$, then the possible elevation lines are $y=0$, $y=1, \ldots, y=j-1$. Summing over $j$ to get what we have currently. Thus we arrive at a combinatorial interpretation 