题目

Problem

The sequence $u_1,u_2,u_3,\ldots$ is defined by

u_{n+1}=1-\frac1{u_n}

u_1=4

(a) Show that this is a periodic sequence of order $3$

(3)

(b) Find the value of

\sum_{n=1}^{180}(5n+3+u_n)

(4)

解答

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从 $u_1=4$ 开始连续代入递推式。如果 $u_4=u_1$ ，而前三项没有更早重复，就说明周期是 $3$ 。

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Given

\begin{align*} u_1=4, \end{align*}

we have

\begin{align*} u_2=&\,1-\frac14=\frac34,\\ u_3=&\,1-\frac{1}{\frac34} =1-\frac43 =-\frac13,\\ u_4=&\,1-\frac{1}{-\frac13} =1+3 =4. \end{align*}

Since $u_4=u_1$ , the sequence repeats:

\begin{align*} 4,\ \frac34,\ -\frac13,\ 4,\ \frac34,\ -\frac13,\ldots \end{align*}

Therefore, this is a periodic sequence of order $3$ .

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把求和拆成两部分：

$\sum(5n+3)$ 是等差数列求和； $\sum u_n$ 每 $3$ 项一组重复， $180$ 项正好是 $60$ 组。

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First,

\begin{align*} \sum_{n=1}^{180}(5n+3) =&\,\frac{180}{2}(8+903)\\ =&\,90(911)\\ =&\,81990. \end{align*}

One cycle of $u_n$ has sum

4+\frac34-\frac13 =\frac{48}{12}+\frac9{12}-\frac4{12} =\frac{53}{12}.

Since

\begin{align*} 180=60\times3, \end{align*}

we get

\begin{align*} \sum_{n=1}^{180}u_n =&\,60\left(\frac{53}{12}\right)\\ =&\,265. \end{align*}

Therefore,

\begin{align*} \sum_{n=1}^{180}(5n+3+u_n) =&\,81990+265\\ =&\,82255. \end{align*}