Inspired by a recent question I tried to generalize the problem and has considered a biased frog which jumps clockwise (left) and counterclockwise (right) with different probabilities $q$ and $p$$(q+p=1)$. The problem appears to be not so simple as it seemed to me initially. Due to the loss of symmetry the expected value of jumps required to expand the set of $m$ visited pads depends on the previous direction of the expansion. Denoting the corresponding values $E_m^-$ and $E_m^+$ for the left and right start positions, respectively, I looked for the value$$E_m=P_m^-E_m^-+P_m^+E_m^+,\tag1$$where $P_m^-$ and $P_m^+$ are the probabilities that the $m-1$-th expansion of the set of visited pads has occurred on the left and on the right ends, respectively.
Knowing $E_m$ the expected number of jumps required to visit all $n$ pads can be easily computed as$$\sum_{m=1}^{n-1}E_m.\tag2$$
After cumbersome calculations (including a guess and its proof by induction) I ended up with a (unexpectedly) nice looking expression:$$E_m(r)=\frac{1+r}{1-r}\left[\frac{(m+1)^2}{1-r^{m+1}}-\frac{2m^2}{1-r^{m}}+\frac{(m-1)^2}{1-r^{m-1}}-1\right],\tag3$$where $r=\frac qp$. One can check that $E_m(1)=\lim_{r\to1}E_m(r)=m$ as in the referred answer. The expression has (as it should) a hidden symmetry $E_m(r)=E_m(1/r)$.
My question: is there a simple or even intuitive way to arrive at the expression $(3)$ ?
