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CIE 9709 2024 June Paper 12 Q5

A Level / CIE / P1

CIE 9709 2024 June Paper 12 Paper · Question 5

题目

Problem

The first and second terms of an arithmetic progression are tan⁡θ\tan\theta and sin⁡θ\sin\theta respectively, where 0<θ<12π0 < \theta < \frac{1}{2}\pi.

(a) Given that θ=14π\theta = \frac{1}{4}\pi, find the exact sum of the first 4040 terms of the progression.

[4]

The first and second terms of a geometric progression are tan⁡θ\tan\theta and sin⁡θ\sin\theta respectively, where 0<θ<12π0 < \theta < \frac{1}{2}\pi.

(b) (i) Find the sum to infinity of the progression in terms of θ\theta.

[2]

(ii) Given that θ=13π\theta = \frac{1}{3}\pi, find the sum of the first 1010 terms of the progression. Give your answer correct to 33 significant figures.

[3]
题目中文翻译

一个等差数列的第一项和第二项分别为 tan⁡θ\tan\theta 和 sin⁡θ\sin\theta，其中 0<θ<12π0 < \theta < \frac{1}{2}\pi。

(a) 已知 θ=14π\theta = \frac{1}{4}\pi，求该数列前 4040 项和的精确值。

[4]

一个等比数列的第一项和第二项分别为 tan⁡θ\tan\theta 和 sin⁡θ\sin\theta，其中 0<θ<12π0 < \theta < \frac{1}{2}\pi。

(b) (i) 求该数列的无穷和，用 θ\theta 表示。

[2]

(ii) 已知 θ=13π\theta = \frac{1}{3}\pi，求该数列前 1010 项的和。答案保留 33 位有效数字。

[3]

解答

(a)

解法一

思路

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先求首项和公差。正弦的精确值要保留到求和结束。

答题过程

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The first term is tan⁡(π/4)=1\tan(\pi/4)=1, and the second is sin⁡(π/4)=2/2\sin(\pi/4)=\sqrt2/2. Hence d=2/2−1d=\sqrt2/2-1, so

S40= 402(2+39(22−1))= 3902−740.\begin{align*} S_{40}=&\,\frac{40}{2}\Big(2+39\Big(\frac{\sqrt2}{2}-1\Big)\Big)\\ =&\,390\sqrt2-740. \end{align*}

(b)(i)

解法一

思路

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等比数列用第二项除以首项求公比。化简后还需检查是否能求无穷和。

答题过程

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The common ratio is sin⁡θ/tan⁡θ=cos⁡θ\sin\theta/\tan\theta=\cos\theta. Since 0<θ<π/20<\theta<\pi/2, we have 0<cos⁡θ<10<\cos\theta<1. The sum to infinity is

tan⁡θ1−cos⁡θ.\frac{\tan\theta}{1-\cos\theta}.

(b)(ii)

解法一

思路

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代入角度后，首项是 3\sqrt3、公比是 1/21/2。这里求的是有限项和。

答题过程

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At θ=π/3\theta=\pi/3, the first term is 3\sqrt3 and the common ratio is 1/21/2. Therefore

S10=3(1−(1/2)10)1−1/2≈3.46S_{10}=\frac{\sqrt3(1-(1/2)^{10})}{1-1/2}\approx3.46

to three significant figures.