Show $f(n)= \frac{ (1-\alpha) a^{n+1} e^{-a}+\alpha b^{n+1} e^{-b} }{ (1-\alpha) a^{n} e^{-a}+\alpha b^{n} e^{-b} }$ is unique for $(\alpha,a,b)$ Suppose we have the following function
\begin{align}
f(n)= \frac{ (1-\alpha) a^{n+1} e^{-a}+\alpha b^{n+1} e^{-b} }{ (1-\alpha) a^{n} e^{-a}+\alpha b^{n} e^{-b} },  \text{ where }n=0,1,2,3,4, \ldots
\end{align}
for some parameters  $\alpha \in (0,1)$, $a>0$ and $b>0$ where $a\neq b$. 
Can we show the function $f$ is always unique for any choice of $\alpha,a$ and $b$?  In other words, if  $f'$ corresponds to the triplet $(\alpha',a',b')$,   then   $f=f'$ if and only if $(\alpha,a,b)=(\alpha',a',b')$.
I tried several examples and think this is true (of course I can be wrong)
What I noticed that think might be helpful is that
if we define
\begin{align}
g(x;a)= e^{-ax} 
\end{align}
then
\begin{align}
f(n)= - \frac{(1-\alpha) g^{(n+1)}(1;a) + \alpha g^{(n+1)}(1;b)}{(1-\alpha) g^{(n)}(1;a) + \alpha g^{(n)}(1;b)}
\end{align}
where $g^{(n)}$ is the $n$-th derivative with respect to $x$. 
Correction:  In the solution, I would like to consider $(\alpha,a,b)$ and $(1-\alpha,b,a)$ to be the same triplet.  The reason is that this result in the change of the order of summation the denominator and numerator stay the same. 
 A: I'm afraid no, we can't. 
If $f$ is the function for a triplet $(\alpha,a,b)$ with $\alpha \ne \frac12$ or $a \ne b$, then the triplet $(1-\alpha,b,a)$ is a different triplet, but $f'=f$ for the corresponding function $f'$.
This can be seen "almost" immediately: If we swap $\alpha$ with $1-\alpha$ and $a$ with $b$, each two summands in both the numerator and denominator just change places.
Also, if $a=b$ then for all $\alpha \in [0,1]$ we have $f(n)=a$ for all $n=0,1,2,3,4, \ldots$
Added:
According to your definition of $\displaystyle{f=f_{\alpha,a,b}}$ together with the restrictions $a \ne b$ and $a \in ]0,1[$, the following is true:
$\displaystyle{f_{\alpha,a,b}=f_{\alpha',a',b'} \Leftrightarrow (\alpha,a,b) = (\alpha',a',b) \lor (\alpha,a,b) = (1-\alpha',b',a')}$.
Note that if $a=b$, the function does not depend on $\alpha$, and if $\alpha=1$ (or $\alpha=0$), the function does not depend on $a$ (or $b$).
Sketch of proof:
With $p:=a+b$ and $q:=ab$, we can prove (by complete induction) that $f(n+1) = p- \frac{q}{f(n)}$.
$\text{Also (seen via calculation), if }a \ne b\text{, then }\alpha\text{ is uniquely determined by }a\text{, }b\text{ and }f(0)\tag{*}$
Now, if we have $f=f'$ with $\displaystyle{f'=f_{\alpha',a',b'}}$, we define $p':=a'+b'$ and $q'=a'b'$ and we find that for $n \ge 0$:
$$p'+\frac{q'}{f(n)}=p+\frac{q}{f(n)} \Leftrightarrow (p'-p)f(n) = q-q'$$
So, $f$ must be constant or $p=p' \land q=q'$. If the latter is true, $a$ and $b$ are the solutions of $x^2-px+q=0$ and so are $a'$ and $b'$, so $\{a,b\}= \{a',b'\}$ which results in $(\alpha,a,b) = (\alpha',a',b) \lor (\alpha,a,b) = (1-\alpha',b',a')$ (because of (*)).
Otherwise, if $f$ is constant, then $f(n)=c$ for all $n$ and so 
$c=p-\frac{q}{c} \Leftrightarrow c^2-pc+q=0 \Leftrightarrow c \in \{a,b\} \Leftrightarrow \alpha \in \{0,1\}$ (again because of (*)).
Remark:
The series $F(n)$ defined by the denominator (and numerator) of $f$, $$F(n):= (1-\alpha)a^n e^{-a}+\alpha b^n e^{-b}$$ is "Fibonacci-like" and obeys to the recursion formula $F(n+1)=p \cdot F(n) - q \cdot F(n-1)$ (with $p=a+b$ and $q=ab$, which does not depend on $\alpha$), and we have $f(n)=\frac{F(n+1)}{F(n)}$.
