3-ча 1. In which of the expressions:

$$
\left(1-x^{2}+x^{3}\right)^{1000}, \quad\left(1+x^{2}-x^{3}\right)^{1000}
$$

after expanding the brackets and combining like terms, is the coefficient of $x^{20}$ greater?
Solve 1. Answer: In the expression $\left(1+x^{2}-x^{3}\right)^{1000}$. Let $P(x)=\left(1-x^{2}+x^{3}\right)^{1000}$ and $Q(x)=\left(1+x^{2}-x^{3}\right)^{1000}$. The coefficient of $x^{20}$ in the polynomial $P(x)$ is the same as in the polynomial $P(-x)=\left(1-x^{2}-x^{3}\right)^{1000}$, and the coefficient of $x^{20}$ in the polynomial $Q(x)$ is the same as in the polynomial $Q(-x)=\left(1+x^{2}+x^{3}\right)^{1000}$. It is clear that the coefficient of $x^{20}$ in the polynomial $\left(1+x^{2}+x^{3}\right)^{1000}$ is greater than in the polynomial $\left(1-x^{2}-x^{3}\right)^{1000}$.

Part 2. Calculate the product with a deficiency to five decimal places (i.e., with an accuracy of 0.00001):

$$
\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right)\left(1-\frac{1}{10^{3}}\right) \cdots\left(1-\frac{1}{10^{99}}\right)
$$

Prove the accuracy of the obtained digits.

Answer 2: 0.89001.

Let $a=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right) \ldots\left(1-\frac{1}{10^{5}}\right)$ and $b=\left(1-\frac{1}{10^{6}}\right) \ldots\left(1-\frac{1}{10^{99}}\right)$. Direct calculations show that

$$
0.89001+\frac{1}{10^{6}}1-x-y$. Therefore, $b>$ $1-\frac{1}{10^{6}}-\frac{1}{10^{7}}-\cdots-\frac{10^{10^{7}}}{10^{99}}>1-\frac{2}{10^{6}}$. Consequently,

$$
\begin{aligned}
& a b0.890011 \cdot 0.999998>0.890009
\end{aligned}
$$