20. A research institute transfers a high-tech technology to a company for production. Both parties agree: the research institute will withdraw 4 million yuan from the company's profit at the end of each year and invest it in other investments, thus earning a profit of 10% annually.
（1）Can the research institute obtain a total income of 130 million yuan after three years?
（2）If the research institute requires a total income of 160 million yuan after three years, how much money should it withdraw from the company's profit at least each year?
20. (1) Let the annual income at the end of each year be $A$ million yuan, with an annual profit of $10\%$, then:
the income from the first year's $A$ million yuan to the $n$-th year is $Q_{1}=A(1+0.1)^{n-1}$;
the income from the second year's $A$ million yuan to the $n$-th year is $Q_{2}=A(1+0.1)^{n-2}$;
the income from the $(n-1)$-th year's $A$ million yuan to the $n$-th year is $Q_{n-1}=A(1+0.1)$;
the income from the $n$-th year is $Q_{n}=A(1+0.1)^{0}=A$.
$Q_{1}, Q_{2} \cdots Q_{n}$ form a geometric sequence. Using the sum formula for a geometric sequence, the total income over these $n$ years is:
$$
Q=Q_{1}+Q_{2}+\cdots+Q_{n-1}+Q_{n}=\sum_{i=1}^{n} A(1+0.1)^{i-1}=\frac{A}{0.1}\left[(1+0.1)^{n}-1\right] \text {. }
$$

When $n=3$, $A=400$, $Q=\frac{400}{0.1} \cdot\left[(1+0.1)^{3}-1\right]=1324$ (million yuan). That is, the income after 3 years is 1324 million yuan, so 1300 million yuan can be obtained after 3 years.
(2) To recover the entire investment within 3 years, i.e., $Q=1600$, then $1600=\frac{A}{0.1}\left[(1+0.1)^{3}-1\right]$, solving for $A \approx 483.4$ (million yuan).
That is, an annual income of at least 483.4 million yuan is required to recover the entire investment within three years.