3B. Determine the first two and the last two digits of the decimal representation of the number $x_{1001}$, if $x_{1}=2$ and $x_{n+1}=\frac{1}{\sqrt[10]{2}} x_{n}+\frac{\sqrt[10]{2}-1}{\sqrt[10]{2}}, n \in \mathbb{N}$.
Solution. From the recursive definition of the sequence, we get

$$
x_{n+1}-1=\frac{1}{\sqrt[10]{2}}\left(x_{n}-1\right), n \in \mathbb{N}
$$

We define the sequence $y_{n}=x_{n}-1, n \in \mathbb{N}$ where $y_{1}=1$ and $y_{n+1}=\frac{1}{\sqrt[10]{2}} y_{n}, n \in \mathbb{N}$. Therefore, $\left\{y_{n}\right\}$ is a geometric progression with the first term $y_{1}=1$ and common ratio $\frac{1}{\sqrt[10]{2}}$. Hence, $y_{n+1}=y_{1}\left(\frac{1}{\sqrt[10]{2}}\right)^{n}, n \in \mathbb{N}$. We have $y_{1001}=\frac{1}{2^{100}}$ and $x_{n}=y_{n}+1, n \in \mathbb{N}$, from which we get

$$
x_{1001}=1+\frac{1}{2^{100}}=1+\frac{5^{100}}{10^{100}}=1+\underbrace{0.00 \ldots 01}_{100 \text { zeros }} \cdot 5^{100}
$$

Thus, the first two digits of the number $x_{1001}$ are 1 and 0, and the last two digits are 2 and 5.