22. Let the real number sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
(2+\mathrm{i})^{n}=a_{n}+b_{n} \mathrm{i}(n \in \mathbf{N}) \text {. }
$$

Then the value of $\sum_{n=0}^{\infty} \frac{a_{n} b_{n}}{7^{n}}$ is ( ).
(A) $\frac{3}{8}$
(B) $\frac{7}{16}$
(C) $\frac{1}{2}$
(D) $\frac{9}{16}$
(E) $\frac{4}{7}$
22. B.
Notice,
$$
\begin{array}{l}
2+\mathrm{i}=\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}} \mathrm{i}\right) \\
=\sqrt{5}(\cos \theta+\mathrm{i} \sin \theta)\left(\theta=\arctan \frac{1}{2}\right). \\
\text { Then }(2+\mathrm{i})^{n}=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}} \mathrm{i}\right)\right)^{n} \\
=(\sqrt{5}(\cos \theta+\mathrm{i} \sin \theta))^{n} \\
=5^{\frac{n}{2}}(\cos n \theta+\mathrm{i} \sin n \theta).
\end{array}
$$

Thus, $(2+\mathrm{i})^{n}=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}} \mathrm{i}\right)\right)^{n}$

Therefore, $a_{n}=5^{\frac{n}{2}} \cos n \theta$,
$$
\begin{array}{l}
b_{n}=5^{\frac{n}{2}} \sin n \theta. \\
\text { Hence } \sum_{n=0}^{\infty} \frac{a_{n} b_{n}}{7^{n}}=\sum_{n=0}^{\infty} \frac{5^{n} \cos n \theta \cdot \sin n \theta}{7^{n}} \\
=\frac{1}{2} \sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^{n} \sin 2 n \theta \\
=\frac{1}{2} \operatorname{Im} \sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^{n} \mathrm{e}^{2 n \theta \mathrm{i}}. \\
\text { Also } \frac{1}{2} \sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^{n} \mathrm{e}^{2 n \theta \mathrm{i}}=\frac{1}{2} \cdot \frac{1}{1-\frac{5}{7} \mathrm{e}^{2 \theta \mathrm{i}}} \\
=\frac{1}{2} \cdot \frac{1}{1-\frac{5}{7}(\cos 2 \theta+\mathrm{i} \sin 2 \theta)} \\
=\frac{1}{2} \cdot \frac{1}{1-\frac{5}{7}\left(\cos ^{2} \theta-\sin ^{2} \theta+2 \mathrm{i} \sin \theta \cdot \cos \theta\right)} \\
=\frac{1}{2} \cdot \frac{1}{1-\frac{5}{7}\left(\frac{4}{5}-\frac{1}{5}+2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} \mathrm{i}\right)} \\
=\frac{1}{2} \cdot \frac{1}{\frac{4}{7}-\frac{4}{7} \mathrm{i}}=\frac{7}{16}+\frac{7}{16} \mathrm{i}, \\
\end{array}
$$

Therefore, $\sum_{n=0}^{\infty} \frac{a_{n} b_{n}}{7^{n}}=\frac{7}{16}$.