15. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}+S_{n}=n$, the sequence $\left\{b_{n}\right\}$ satisfies $b_{1}=a_{1}$, and $b_{n}=a_{n}-$ $a_{n-1}(n \geqslant 2)$.
(1) Find the general term formula for $\left\{a_{n}\right\}$;
(2) Find the general term formula for $\left\{b_{n}\right\}$;
(3) Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $B_{n}$.
15. (1) Since $a_{n}+S_{n}=n, a_{n-1}+S_{n-1}=n-1$, subtracting the two equations gives $2 a_{n}-a_{n-1}=1$, then $2\left(a_{n}-\right.$ $1)=\left(a_{n-1}-1\right)$, hence $\left\{a_{n}-1\right\}$ is a geometric sequence, so $a_{n}-1=\left(a_{1}-1\right) \cdot\left(\frac{1}{2}\right)^{n-1}$. And $a_{1}+S_{1}=1$, thus $a_{1}=\frac{1}{2}$, therefore $a_{n}-1=-\left(\frac{1}{2}\right)^{n}$, which means $a_{n}=1-\left(\frac{1}{2}\right)^{n}$;
(2) $b_{n}=a_{n}-a_{n-1}=\left(\frac{1}{2}\right)^{n-1}-\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n}(n \geqslant 2)$, and $b_{1}=a_{1}=\frac{1}{2}$, hence $b_{n}=$ $\left(\frac{1}{2}\right)^{n}(n \geqslant 1)$
(3) $b_{1}+b_{2}+\cdots+b_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=a_{n}=1-\left(\frac{1}{2}\right)^{n}=B_{n}$.