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 $k^2 - 7k - 60 = 0$

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

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

(9)

题目中文翻译

某软件开发者发布了一款应用。发布后前三个月的下载量（单位：千次）为 $2k-15$ 、 $k$ 、 $k+4$ ，其中 $k$ 为常数。

已知每月下载量构成等比数列，

(a) 证明 $k^2-7k-60=0$

(b) 预测第4个月的下载量

该应用的累计下载量预计在第 $N$ 个月首次超过300万次。

（9分）

解答

等比数列相邻项之比相等： $\dfrac{k}{2k-15} = \dfrac{k+4}{k}$ 。解二次方程得 $k$ ，取正值。求公比 $r$ ，第4个月 $= a \cdot r^3$ 。前 $N$ 项和 $S_N = a\dfrac{r^N - 1}{r - 1} > 3000$ 。

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

k^2 =&\, (2k - 15)(k + 4) \\[4mm] k^2 =&\, 2k^2 + 8k - 15k - 60 \\[4mm] k^2 =&\, 2k^2 - 7k - 60 \\[4mm] k^2 - 7k - 60 =&\, 0 \end{align*}$$ $$\boxed{k^2 - 7k - 60 = 0}$$ **(b)** Solving $(k - 12)(k + 5) = 0$: $$k = 12 \quad \text{or} \quad k = -5$$ Since downloads must be positive, $2k - 15 > 0 \implies k > 7.5$, so $k = 12$. The three terms are: $9,\; 12,\; 16$. Common ratio: $r = \dfrac{12}{9} = \dfrac{4}{3}$. 4th month downloads $= 16 \times \dfrac{4}{3} = \dfrac{64}{3} \approx 21.33$ thousand. $$\boxed{\dfrac{64}{3} \text{ thousand} \approx 21{,}333}$$ **(c)** $a = 9$, $r = \dfrac{4}{3}$, and downloads are in thousands, so we need $S_N > 3000$. $$\begin{align*} S_N =&\, 9 \times \frac{\left(\frac{4}{3}\right)^N - 1}{\frac{4}{3} - 1} \\[4mm] =&\, 9 \times \frac{\left(\frac{4}{3}\right)^N - 1}{\frac{1}{3}} \\[4mm] =&\, 27\left[\left(\frac{4}{3}\right)^N - 1\right] \end{align*}$$ $$\begin{align*} 27\left[\left(\frac{4}{3}\right)^N - 1\right] >&\, 3000 \\[4mm] \left(\frac{4}{3}\right)^N - 1 >&\, \frac{3000}{27} = \frac{1000}{9} \\[4mm] \left(\frac{4}{3}\right)^N >&\, \frac{1009}{9} \\[4mm] N \ln\frac{4}{3} >&\, \ln\frac{1009}{9} \\[4mm] N >&\, \frac{\ln(1009/9)}{\ln(4/3)} \approx \frac{4.719}{0.2877} \approx 16.4 \end{align*}$$ Since $N$ must be an integer: $$\boxed{N = 17}$$