正数数列 \(\{a_n\}\) 满足 \(S_n + 1 = \dfrac{S_n+4}{2a_n - S_n}\), 其中 \(S_n\) 是 \(\{a_n\}\) 的前 \(n\) 项和, 求 \(\{a_n\}\) 的通项公式.
分析和解答
不难计算得到 \(a_1 = S_1 = 2\). 另根据已知条件, \(S_n + 1 = \dfrac{S_n+4}{S_n - 2S_{n-1}}\), 可变形得
\[
S_{n-1} = \dfrac{S_n^2 - 4}{2(S_n+1)}
\]
直接处理数列 \(\{S_n\}\), 其不动点方程为 \(x = \dfrac{x^2-4}{2(x+1)}\), 解得 \(x = -1 \pm \sqrt3 i = 2 e^{i \cdot \frac{\pm 2\pi}3}\). 记 \(\omega_1 = e^{i \cdot \frac{2\pi}3}\), \(\omega_2 = e^{i \cdot \frac{-2\pi}3}\), 则不动点方程的两根分别为 \(2\omega_1\) 和 \(2\omega_2\).
\(\omega_{1,2}\) 为三次单位根, 有
- \(\omega_1^2 + \omega_1 + 1 = \omega_2^2 + \omega_2 + 1 = 0\);
- \(\omega_1 = \omega_2^2\), \(\omega_2 = \omega_1^2\);
- \(\omega_1 = \overline{\omega_2}\)
分别用两个不动点处理数列 \(\{S_n\}\) 的递推关系, 有
\[
\begin{aligned}
S_{n-1} - 2 \omega_1 &= \dfrac{(S_n - 2 \omega_1)^2}{2(S_n+1)} \\
S_{n-1} - 2 \omega_2 &= \dfrac{(S_n - 2 \omega_2)^2}{2(S_n+1)}
\end{aligned}
\]
两式相除得
\[
\dfrac{S_{n-1} - 2 \omega_1}{S_{n-1} - 2 \omega_2} = \left( \dfrac{S_n - 2 \omega_1}{S_n - 2 \omega_2} \right)^2
\]
比较两边的模和辐角主值. 由于 \(\forall n \in \mathbb N^*, S_n \in \mathbb R\), 故 \(S_n - 2\omega_1 = \overline{S_n - 2\omega_2}\), 于是 \(\left| \dfrac{S_n - 2 \omega_1}{S_n - 2 \omega_2} \right| = 1\), 模的条件自然满足.
设 \(S_n - 2\omega_1\) 的辐角主值为 \(\theta_n\), 则 \(S_n - 2\omega_2\) 的辐角主值为 \(-\theta_n\), 于是
\[
\dfrac{S_n - 2 \omega_1}{S_n - 2 \omega_2} = e^{i \cdot 2\theta_n}
\]
因此
\[
e^{i \cdot 2\theta_{n-1}} = e^{i \cdot 4\theta_n}, \quad 2\theta_{n-1} = 4\theta_n + 2k\pi
\]
由于 \(\{a_n\}\) 是正数数列, \(\{S_n\}\) 也是正数数列且单调递增, 故复数 \(S_n - 2\omega_1\) 的实部 \(\Re(S_n - 2\omega_1) = S_n+1\) 恒为正且单调递增, 而虚部 \(\Im (S_n - 2\omega_1) = -\sqrt3\) 为负数常值, 故 \(\{\theta_n\}\) 都是第四象限角且单调递增, 因而只能是
\[
2\theta_{n-1} = 4\theta_n, \quad \theta_n = \dfrac1{2^{n-1}} \cdot \theta_1
\]
其中, \(\theta_1\) 是 \(S_1 - 2 \omega_1 = 2 - 2 \omega_1 = 3-\sqrt3i\) 的辐角, \(\theta_1 = -\dfrac\pi6\). 于是 \(\theta_n = -\dfrac\pi{3\cdot 2^n}\).
由 \(\dfrac{S_n - 2 \omega_1}{S_n - 2 \omega_2} = e^{i \cdot 2\theta_n}\) 反推得
\[
\begin{aligned}
S_n &= \dfrac{2 (\omega_1 - \omega_2 \cdot e^{i \cdot 2\theta_n})}{1 - e^{i \cdot 2\theta_n}} \\
&= 2\cdot \dfrac{(\omega_1 - \omega_2 \cdot e^{i \cdot 2\theta_n})(1 - e^{-i \cdot 2\theta_n})}{(1 - e^{i \cdot 2\theta_n})(1 - e^{-i \cdot 2\theta_n})} \\
&= 2\cdot \dfrac{\omega_1 + \omega_2 - \omega_1 \cdot e^{-i \cdot 2\theta_n} - \omega_2 \cdot e^{i \cdot 2\theta_n}}{1 - e^{i \cdot 2\theta_n} - e^{-i \cdot 2\theta_n} + 1} \\
&= 2\cdot \dfrac{ -1 -2 \cos \left( \dfrac{2\pi}3 - 2\theta_n \right)}{2 - 2\cos2\theta_n} \\
&= \dfrac{-1+\cos2\theta_n - \sqrt3 \sin2\theta_n}{1-\cos2\theta_n} \\
&= \dfrac{-\sqrt3 \sin2\theta_n}{1-\cos2\theta_n} -1 \\
&= -\sqrt3 \cot \theta_n - 1 \\
&= \sqrt3 \cot \dfrac\pi{3\cdot 2^n} - 1
\end{aligned}
\]
注意到 \(S_0 = \sqrt3 \cot \dfrac\pi3 - 1 = 0\), 有 \(\forall n \in \mathbb N^*\), 都有
\[
\begin{aligned}
a_n &= S_n - S_{n-1} \\
&= \sqrt3 \left( \cot \dfrac\pi{3\cdot 2^n} - \cot \dfrac\pi{3\cdot 2^{n-1}} \right)
= \dfrac{\sqrt3}{\sin\dfrac\pi{3\cdot2^{n-1}}}
\end{aligned}
\]
注意到
\[
\begin{aligned}
\cot \theta - \cot 2\theta &= \dfrac{\cos\theta}{\sin\theta} - \dfrac{\cos2\theta}{\sin2\theta} \\
&= \dfrac{\sin2\theta\cos\theta - \cos2\theta\sin\theta}{\sin\theta \sin2\theta} \\
&= \dfrac{\sin\theta}{\sin\theta \sin2\theta} = \dfrac1{\sin2\theta}
\end{aligned}
\]
