3. Expand $(1+0.2)^{1000}$ using the binomial theorem, i.e. let $C_{1000}^{0}(0.2)^{0}+C_{1000}^{1}(0.2)^{1}+\cdots+$ $C_{1000}(0.2)^{1000}=A_{0}+A_{1}+\cdots+A_{1000}$. Let $A_{h}=C_{1000}^{k}(0.2)^{h}, k=0,1,2, \cdots$, 1000 . Find the value of $k$ for which $A_{k}$ is the largest.
3. For $1 \leqslant k \leqslant 1000$,
$$
\begin{array}{l}
\frac{A_{\mathrm{s}}}{A_{\mathrm{k}-1}}=\frac{\frac{1000!}{k!(1000-k)!}(0.2)^{k}}{\frac{1000!}{(k-1)!(1001-k)!}(0.2)^{k}} \\
=\frac{1001-k}{k}(0.2) \text {. } \\
\end{array}
$$

When and only when $1001-k>5k$, this ratio is greater than 1. That is, when $k \leqslant 166$, the above ratio is greater than 1.
Therefore, $A_{0}A_{108}>\cdots>A_{1000 \text { . }}$
And $A_{1 \in \theta}>A_{107}$,
which means when $\dot{x}=1$, $A$ is the largest.