1. Consider the sequence $\left(d_{n}\right)_{n \geq 1}$ defined descriptively as follows:

$$
d_{1}=0.1 ; d_{2}=0.11 ; d_{3}=0.111 ; \ldots ; d_{n}=0, \underbrace{11 \ldots 1}_{n \text { times }}
$$

a) Find the first 2016 decimal places of the number $\sqrt{d_{4032}}$.

b) Find the sum: $\sum_{k=1}^{n} 10^{k} \cdot d_{k}$.
1.a) Since $x \in(0,1) \Rightarrow x<\sqrt{x}<1(*)$. We consider

$x=0, \underbrace{99 \ldots 9}_{4032 \text { times }}=1-\frac{1}{10^{4032}}$. Using the relation $\left({ }^{*}\right)$, we deduce that the first 2016 decimal places of $\sqrt{x}$ will be equal to 9. Since $d_{4032}=\frac{x}{9}, \sqrt{d_{4032}}=\frac{\sqrt{x}}{3}=0, \underbrace{33 \ldots 3}_{2016 \text { times }} \ldots$. In conclusion, the first 2016 decimal places of the number $\sqrt{d_{4032}}$ are equal to 3.
b) $\sum_{k=1}^{n} 10^{k} \cdot d_{k}=1+11+111+\cdots+\underbrace{111 \ldots 1}_{n \text { times }}=$

$=\frac{1}{9} \cdot(10-1)+\frac{1}{9} \cdot\left(10^{2}-1\right)+\ldots+\frac{1}{9} \cdot\left(10^{n}-1\right)=\frac{1}{9} \cdot\left(\frac{10 \cdot\left(10^{n}-1\right)}{10-1}-n\right)=$ $=\frac{10}{81} \cdot\left(10^{n}-1\right)-\frac{n}{9}$.