6 Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation $a_{n+1}=2 a_{n}+2^{n}-1(n \in$
$\left.\mathbf{N}^{*}\right)$, and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence, then the value of $\lambda$ is $\qquad$.
$6-1$ Hint: Since
$$
\begin{aligned}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} & =\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
& =\frac{2^{n}-1-\lambda}{2^{n+1}},
\end{aligned}
$$

from $\frac{2^{n}-1-\lambda}{2^{n+1}}$ being a constant, we know $\lambda=-1$.