6. Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$
a_{n+1}=2 a_{n}+2^{n}-1\left(n \in \mathbf{N}_{+}\right) \text {, }
$$

and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence. Then the value of $\lambda$ is $\qquad$
6. -1 .

Notice,
$$
\begin{array}{l}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2^{n}-1-\lambda}{2^{n+1}} . \\
\text { From } \frac{2^{n}-1-\lambda}{2^{n+1}} \text { being a constant, we know } \lambda=-1 .
\end{array}
$$