2. Let $x_{1}=\frac{1}{4}, x_{2}=\frac{3}{16}, x_{n+2}=\frac{3}{16}-\frac{1}{4}\left(x_{1}\right.$ $\left.+x_{2}+\cdots+x_{n}\right)(n=1,2, \cdots)$. Then $x_{n}=$
2. $\frac{n+1}{2^{n+2}}$.

From the given, we have $x_{n+2}-x_{n+1}=-\frac{1}{4} x_{n}$, which means $x_{n+2}=x_{n+1}-\frac{1}{4} x_{n}$.
Let $a_{n}=2^{n} x_{n}$, then
$$
a_{1}=\frac{1}{2}, a_{2}=\frac{3}{4}, a_{n+2}=2 a_{n+1}-a_{n} \text {. }
$$

Then $a_{n+2}-a_{n+1}=a_{n+1}-a_{n}=\cdots=a_{2}-a_{1}=\frac{1}{4}$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term $a_{1}=\frac{1}{2}$ and common difference $d=\frac{1}{4}$.
Hence, $a_{n}=a_{1}+(n-1) d=\frac{1}{2}+\frac{1}{4}(n-1)=\frac{n+1}{4}$.
Thus, $x_{n}=\frac{a_{n}}{2^{n}}=\frac{n+1}{2^{n+2}}$.