2. (2 points) The sequence $x_{n}$ is defined by the following conditions: $x_{1}=3, x_{1}+x_{2}+\ldots+$ $x_{n-1}+\frac{4}{3} x_{n}=4$. Find $x_{2015}$.

Answer: $x_{2015}=\frac{3}{4^{2014}}$.
Solution:

Notice that the equality $x_{1}+x_{2}+\ldots+x_{n-1}+\frac{4}{3} x_{n}=4$ for $n=1$ also holds: $\frac{4}{3} x_{1}=4$.

For $n \geqslant 2$, by subtracting the equality $x_{1}+x_{2}+\ldots+\frac{4}{3} x_{n-1}=4$ from the equality $x_{1}+x_{2}+\ldots+x_{n-1}+\frac{4}{3} x_{n}=4$, we get that $\frac{4}{3} x_{n}-\frac{1}{3} x_{n-1}=0$, from which $x_{n}=\frac{1}{4} x_{n-1}$ for $n \geqslant 2$.

Thus, $x_{n}$ is a geometric progression with the first term 3 and the common ratio $\frac{1}{4}$, so $x_{2015}=\frac{3}{4^{2014}}$.