Need help with non-recursive definition So I'm trying to find a non-recursive definition for $b_n$.   I'm given $$b_0=1$$ $$b_{n+1}=2b_n-1$$
Does this mean I'm trying to find a number for $b_n$ that fits that algorithm?
Update:
Proof by induction.  Let P(n) be that for any $n$, $b_n=1$.
As our base case, we prove P(0), that $b_0=1$, which is obvious from the problem.
For our inductive step, assume for some n ∈ N that P(n) holds.  We prove that P(n+1) holds. 
$$b_{n+1}=2b_n-1$$
$$b_n=(b_{n+1}+1)/2$$
I'm not sure what to do next here.

Use Zev's hint to gain understanding about what you need to prove. Often times you will have a crazy nested expression to help find a closed form of the sequence. This is common strategy.
Here is a formal proof if you need some help/verification.
Let $(b_n)$ be the sequence defined by$$
 b_{n} =\left\{
        \begin{array}{ll}
            1 & \quad n = 0 \\
            2b_{n-1}-1 & \quad n > 0
        \end{array}
    \right.
$$
Let $P(n)$ be the statement: $1 = b_n$ for all $n\geq 0$.
$\bf Base:$ Let $n=0$. The left hand side of the equation is always $1$ and the right hand side is $b_0 = 1$ by definition of $b_n$. Thus $P(0)$ is true.
$\bf Step:$ Assume $P(n)$ is true. That is $1 = b_n$ for $n\geq0$.
Consider $b_{n+1}$ [We must show that $b_{n+1}=1$]
\begin{align}
b_{n+1} =& 2b_n -1 &\quad \text{by definition of $b_n$}\\
=&2*1 -1 &\quad \text{by the induction hypothesis} \\
=& 1
\end{align}
Thus $P(n) \implies P(n+1)$, and therefore by the principle of mathematical induction $1 = b_n$ for all $n \geq 0$
