题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

A software developer released an app to download.

The numbers of downloads of the app each month, in thousands, for the first three months after the app was released were

2k - 15, \quad k, \quad k + 4

where $k$ is a constant.

Given that the numbers of downloads each month are modelled as a geometric series,

(a) show that $k^2 - 7k - 60 = 0$

(2)

(b) predict the number of downloads in the 4th month.

(4)

The total number of all downloads of the app is predicted to exceed 3 million for the first time in the $N$ th month.

(3)

题目中文翻译

本题必须展示全部计算步骤。完全依赖计算器技术的解答不予接受。

一名软件开发者发布了一款可供下载的应用。发布后前三个月的下载量（以千次为单位）依次为 $2k-15$ 、 $k$ 、 $k+4$ ，其中 $k$ 为常数。

已知每月下载量用等比数列建模，

(a) 证明 $k^2-7k-60=0$ 。（2 分）

(b) 预测第四个月的下载量。（4 分）

预计累计下载量在第 $N$ 个月首次超过 300 万次。

解答

(a)

解法一

思路

展开

等比数列相邻项之比相等。列出比例并去分母、展开，逐步整理出给定的二次方程。

答题过程

展开

For a geometric series, $\dfrac{k}{2k-15} = \dfrac{k+4}{k}$ .

\begin{align*} k^2 =&\, (2k - 15)(k + 4) \\ k^2 =&\, 2k^2 + 8k - 15k - 60 \\ k^2 =&\, 2k^2 - 7k - 60 \\ k^2 - 7k - 60 =&\, 0 \end{align*}

\boxed{k^2 - 7k - 60 = 0}

(b)

解法一

思路

展开

承接 (a) 解出 $k$ ，以下载量为正排除负根；求公比，再将第三个月乘一次公比。最后把千次换成次数。

答题过程

展开

Solving $(k - 12)(k + 5) = 0$ :

k = 12 \quad \text{or} \quad k = -5

Since downloads must be positive, $2k - 15 > 0 \implies k > 7.5$ , so $k = 12$ .

The three terms are: $9,\; 12,\; 16$ .

Common ratio: $r = \dfrac{12}{9} = \dfrac{4}{3}$ .

The predicted fourth-month downloads are

\begin{align*} u_4=&\,16\times\frac43=\frac{64}{3}\text{ thousand}\\ =&\,\frac{64000}{3}\text{ downloads}\\ \approx&\,\boxed{21300}\text{ downloads (3 s.f.)}. \end{align*}

用首项公式也可验证： $u_4=9(4/3)^3=64/3$ 千次，与第三项乘一次公比相同。

(c)

解法一

思路

展开

用前面得到的首项 9 与公比 $4/3$ 列累计下载量的不等式。单位为千次，300 万对应 3000；取对数并选择超过界限的最小整数。

答题过程

展开

$a = 9$ , $r = \dfrac{4}{3}$ , and downloads are in thousands, so we need $S_N > 3000$ .

\begin{align*} S_N =&\, 9 \times \frac{\Big(\frac{4}{3}\Big)^N - 1}{\frac{4}{3} - 1} \\ =&\, 9 \times \frac{\Big(\frac{4}{3}\Big)^N - 1}{\frac{1}{3}} \\ =&\, 27\Big[\Big(\frac{4}{3}\Big)^N - 1\Big] \end{align*}

\begin{align*} 27\Big[\Big(\frac{4}{3}\Big)^N - 1\Big] >&\, 3000 \\ \Big(\frac{4}{3}\Big)^N - 1 >&\, \frac{3000}{27} = \frac{1000}{9} \\ \Big(\frac{4}{3}\Big)^N >&\, \frac{1009}{9} \\ N \ln\frac{4}{3} >&\, \ln\frac{1009}{9} \\ N >&\, \frac{\ln(1009/9)}{\ln(4/3)} \\ N >&\,16.405229574\ldots \end{align*}

Since $N$ must be an integer:

\boxed{N = 17}

P2 2023 May Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程