All five letter words are made using all the letters A, B, C, D, E...

All five letter words are made using all the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $\mathrm{P}$\left(\mathrm{W}_{\mathrm{n}}\right)$of choosing the word$\mathrm{W}_{\mathrm{n}}$satisfy$\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n} \gt 1$.<br/>If$\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}$, then$\alpha+\beta is equal to : _______