Possible Duplicate:
Solving a Recurrence Relation/Equation, is there more than 1 way to solve this?
I am trying to solve following recurrence relation
$a(n)=2a(n-1)+1\;.$
I have divided both side by $2^n$, so get $a(n)2^{-n}=2^{1-n}a(n-1)+2^{-n}\;.$
After I put $n=1$ I have got $a(1)=2a(0)$, so $a(0)=1/2$, but how to continue for the general solution? I can't use formula of quadratic equation, namely $k^2-2k-1=0$, because in this case $k_1=1+\sqrt2$ and $k_2=1-\sqrt2$, but it does not help me to find actual solution.