题目

Problem

A geometric series has first term $a$ and common ratio $r$ .

Given that

(a) show that

r=-\frac12

(2)

(b) find the sum to infinity of the series.

(4)

解答

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第 $3$ 项是 $ar^2$ ，第 $6$ 项是 $ar^5$ 。两式相除可以消去 $a$ ，得到 $r^3$ 。

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The third term is

\begin{align*} ar^2=64. \end{align*}

The sixth term is

\begin{align*} ar^5=-8. \end{align*}

Divide the second equation by the first:

\begin{align*} \frac{ar^5}{ar^2}=&\,\frac{-8}{64}\\ r^3=&\,-\frac18\\ r=&\,-\frac12. \end{align*}

This is the required result.

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先用 $ar^2=64$ 和 $r=-\frac12$ 求出首项 $a$ ，再用无穷等比级数公式。

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Using $ar^2=64$ and $r=-\frac12$ ,

\begin{align*} a\left(-\frac12\right)^2=&\,64\\ \frac a4=&\,64\\ a=&\,256. \end{align*}

Since $|r|<1$ , the sum to infinity is

\begin{align*} S_\infty=&\,\frac{a}{1-r}\\ =&\,\frac{256}{1-\left(-\frac12\right)}\\ =&\,\frac{256}{\frac32}\\ =&\,\frac{512}{3}. \end{align*}