## Task A-4.1.

Let $\left(x_{n}\right)$ be a sequence such that $x_{0}=1, x_{1}=2$, with the property that the sequence $\left(y_{n}\right)$ is given by the relation

$$
y_{n}=\binom{n}{0} x_{0}+\binom{n}{1} x_{1}+\cdots+\binom{n}{n} x_{n}, \quad \text { for } n \in \mathbb{N}_{0}
$$

is a geometric sequence. Determine $x_{2020}$.
## Solution.

By substituting into the given equation, we get $y_{0}=1$, and $y_{1}=1+2=3$. Since $\left(y_{n}\right)$ is a geometric sequence, we conclude that $y_{n}=3^{n}$ for all $n \in \mathbb{N}_{0}$.

We will prove by mathematical induction that $x_{n}=2^{n}$ for all $n \in \mathbb{N}_{0}$.

For $n=0$ and $n=1$, we have $x_{0}=1=2^{0}$ and $x_{1}=2=2^{1}$, so the base case of induction is satisfied.

Assume that for all $k \leqslant n$ it holds that $x_{k}=2^{k}$. We will prove that $x_{n+1}=2^{n+1}$. According to the induction hypothesis, we have

$$
\begin{aligned}
& y_{n+1}=\binom{n+1}{0} x_{0}+\binom{n+1}{1} x_{1}+\cdots+\binom{n+1}{n} x_{n}+\binom{n+1}{n+1} x_{n+1} \\
&=\binom{n+1}{0} 2^{0}+\binom{n+1}{1} 2^{1}+\cdots+\binom{n+1}{n} 2^{n}+\binom{n+1}{n+1} x_{n+1} \\
&=\binom{n+1}{0} 2^{0}+\binom{n+1}{1} 2^{1}+\cdots+\binom{n+1}{n} 2^{n}+\binom{n+1}{n+1} 2^{n+1} \\
& \quad-\binom{n+1}{n+1}\left(2^{n+1}-x_{n+1}\right) \\
&=(1+2)^{n+1}-2^{n+1}+x_{n+1} \\
&=3^{n+1}-2^{n+1}+x_{n+1}
\end{aligned}
$$

where the second-to-last equality holds by the binomial theorem. Since $y_{n+1}=3^{n+1}$, it follows that $x_{n+1}=2^{n+1}$, thus completing the induction step.

Therefore, for every $n \in \mathbb{N}_{0}$, we have $x_{n}=2^{n}$, and in particular, $x_{2020}=2^{2020}$.