AMC12 2020 A
AMC12 2020 A · Q22
AMC12 2020 A · Q22. It mainly tests Functions basics, Complex numbers (rare).
Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that $(2 + i)^n = a_n + b_n i$ for all integers $n \geq 0$, where $i = \sqrt{-1}$. What is $\sum_{n=0}^\infty \frac{a_n b_n}{7^n}$?
设$(a_n)$和$(b_n)$是实数序列,使得$(2 + i)^n = a_n + b_n i$对所有整数$n \geq 0$成立,其中$i = \sqrt{-1}$。求$\sum_{n=0}^\infty \frac{a_n b_n}{7^n}$?
(A)
$\frac{3}{8}$
$\frac{3}{8}$
(B)
$\frac{7}{16}$
$\frac{7}{16}$
(C)
$\frac{1}{2}$
$\frac{1}{2}$
(D)
$\frac{9}{16}$
$\frac{9}{16}$
(E)
$\frac{4}{7}$
$\frac{4}{7}$
Answer
Correct choice: (B)
正确答案:(B)
Solution
Answer (B): It follows from the original equality that
$$\left(\frac{2+i}{\sqrt{7}}\right)^{2n}=\frac{(a_n+b_ni)^2}{7^n}=\left(\frac{a_n^2-b_n^2}{7^n}\right)+\left(\frac{2a_nb_n}{7^n}\right)i.$$
Therefore
$$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{2+i}{\sqrt{7}}\right)^{2n}\right)$$
$$=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{3+4i}{7}\right)^n\right)$$
$$=\frac12\,\mathrm{Im}\left(\frac{1}{1-\frac{3+4i}{7}}\right).$$
The preceding geometric series converges as shown because the modulus of the common ratio is less than 1. From
$$\frac{1}{1-\frac{3+4i}{7}}=\frac{7}{4-4i}=\frac14\left(\frac{7}{1-i}\cdot\frac{1+i}{1+i}\right)=\frac78(1+i),$$
it follows that
$$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac{7}{16}.$$
答案(B):由原等式可得
$$\left(\frac{2+i}{\sqrt{7}}\right)^{2n}=\frac{(a_n+b_ni)^2}{7^n}=\left(\frac{a_n^2-b_n^2}{7^n}\right)+\left(\frac{2a_nb_n}{7^n}\right)i.$$
因此
$$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{2+i}{\sqrt{7}}\right)^{2n}\right)$$
$$=\frac12\,\mathrm{Im}\left(\sum_{n=0}^{\infty}\left(\frac{3+4i}{7}\right)^n\right)$$
$$=\frac12\,\mathrm{Im}\left(\frac{1}{1-\frac{3+4i}{7}}\right).$$
上述等比级数收敛,因为公比的模小于 1。并且
$$\frac{1}{1-\frac{3+4i}{7}}=\frac{7}{4-4i}=\frac14\left(\frac{7}{1-i}\cdot\frac{1+i}{1+i}\right)=\frac78(1+i),$$
从而
$$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac{7}{16}.$$
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