2. Let $x_{0}=1990$ and $x_{n}=-\frac{1990}{n}\left(x_{0}+x_{1}+\ldots+x_{n-1}\right)$, for $n \geq 1$. Calculate the sum

$$
x_{0}+2 x_{1}+2^{2} x_{2}+\ldots+2^{1990} x_{1990}
$$
Solution. Od

\[
x_{0}+x_{1}+\ldots+x_{n-1}=-\frac{n}{1990} x_{n}
\]

provo za \( n=k \), a potom za \( n=k-1 \) dobivame

\[
\begin{aligned}
x_{k} & =-\frac{1990}{k}\left(x_{0}+x_{1}+\ldots+x_{k-2}+x_{k-1}\right)=-\frac{1990}{k}\left(-\frac{1990}{k-1} x_{k-1}+x_{k-1}\right) \\
& =-\frac{1990-k+1}{k} x_{k-1}=\ldots=(-1)^{k}\binom{1990}{k} x_{0} .
\end{aligned}
\]

Spored toa,

\[
\sum_{k=0}^{1990} 2^{k} x_{k}=x_{0} \sum_{k=0}^{1990} 2^{k}(-1)^{k}\binom{1990}{k}=(1-2)^{1990} x_{0}=x_{0}=1990
\]

---

Translation:

Solution. From

\[
x_{0}+x_{1}+\ldots+x_{n-1}=-\frac{n}{1990} x_{n}
\]

we prove for \( n=k \), and then for \( n=k-1 \) we get

\[
\begin{aligned}
x_{k} & =-\frac{1990}{k}\left(x_{0}+x_{1}+\ldots+x_{k-2}+x_{k-1}\right)=-\frac{1990}{k}\left(-\frac{1990}{k-1} x_{k-1}+x_{k-1}\right) \\
& =-\frac{1990-k+1}{k} x_{k-1}=\ldots=(-1)^{k}\binom{1990}{k} x_{0} .
\end{aligned}
\]

According to this,

\[
\sum_{k=0}^{1990} 2^{k} x_{k}=x_{0} \sum_{k=0}^{1990} 2^{k}(-1)^{k}\binom{1990}{k}=(1-2)^{1990} x_{0}=x_{0}=1990
\]