Choosing the drift that makes an exponential walk fair

Let $X_1,X_2,\dots$ be independent, each equal to $+1$ or $-1$ with probability $\tfrac12$, and let $c$ be a constant. Start from $M_0=1$ and for $n\ge1$ put $M_n=e^{X_n+c}\,M_{n-1}$. Exactly one value of $c$ makes $(M_n)$ a martingale for the filtration generated by the $X_i$. Writing that value as $c=-\log\left(ae^b+ce^d\right)$ (natural log), compute $a+b+c+d$.

$1$. We need $\mathbb{E}[e^{X_n+c}]=1$, i.e. $e^{c}\cdot\tfrac12(e+e^{-1})=1$, so $c=-\log\left(\tfrac12e^{1}+\tfrac12e^{-1}\right)$. Thus $a=c=\tfrac12$, $b=1$, $d=-1$, and the sum is $\tfrac12+1+\tfrac12-1=1$.