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CIE 9231 2024 June Paper 23 Q4

A Level / CIE / FP2

CIE 9231 2024 June Paper 23 Paper · Question 4

题目

Problem

The diagram shows the curve with equation y=x−2y=x^{-2} for 2≤x≤N2\le x\le N together with a set of (N−2)(N-2) rectangles of unit width.

(a) By considering the sum of the areas of these rectangles, show that

∑r=1N1r2>32−1N+1N2.\sum_{r=1}^{N}\frac{1}{r^2}>\frac{3}{2}-\frac{1}{N}+\frac{1}{N^2}.
[5]

(b) Use a similar method to find, in terms of NN, an upper bound for

∑r=1N1r2.\sum_{r=1}^{N}\frac{1}{r^2}.
[3]

(c) Deduce lower and upper bounds for

∑r=1∞1r2.\sum_{r=1}^{\infty}\frac{1}{r^2}.
[2]
题目中文翻译

图中显示曲线 y=x−2y=x^{-2},其中 2≤x≤N2\le x\le N,以及一组宽度均为 1 的 N−2N-2 个矩形。

(a) 通过考虑这些矩形的面积之和,证明

∑r=1N1r2>32−1N+1N2\sum_{r=1}^{N}\frac{1}{r^2}>\frac{3}{2}-\frac{1}{N}+\frac{1}{N^2}
(5)

(b) 使用类似方法,求关于 NN 的下列和式的上界:

∑r=1N1r2\sum_{r=1}^{N}\frac{1}{r^2}
(3)

(c) 推出

∑r=1∞1r2\sum_{r=1}^{\infty}\frac{1}{r^2}

的下界和上界。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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x−2x^{-2} 递减,题图的左端点矩形在曲线上方。面积和从 22 加到 N−1N-1,再补上首尾两项得到题目要求的和。

答题过程

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For N>2N>2, the left-endpoint rectangles of width one lie above f(x)=x−2f(x)=x^{-2} on the interior of each interval. The left-endpoint areas are 2−22^{-2}, 3−23^{-2}, 4−24^{-2}, …\ldots, (N−2)−2(N-2)^{-2}, (N−1)−2(N-1)^{-2}. Hence

∑r=2N−11r2> ∫2Nx−2 dx= [−x−1]2N=12−1N.\begin{align*} \sum_{r=2}^{N-1}\frac1{r^2}>&\,\int_2^N x^{-2}\,\mathrm{d}x\\ =&\,[-x^{-1}]_2^N=\frac12-\frac1N. \end{align*}

Adding 11 and 1/N21/N^2 to both sides gives

∑r=1N1r2>32−1N+1N2.\sum_{r=1}^N\frac1{r^2}>\frac32-\frac1N+\frac1{N^2}.

(b)

解法一

思路

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换用右端点,矩形面积和小于积分。这个和从 33 加到 NN,补上 r=1,2r=1,2 两项即可。

答题过程

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The right-endpoint areas are 3−23^{-2}, 4−24^{-2}, 5−25^{-2}, …\ldots, (N−1)−2(N-1)^{-2}, N−2N^{-2}. These rectangles lie below the decreasing curve, so

∑r=3N1r2< ∫2Nx−2 dx= 12−1N.\begin{align*} \sum_{r=3}^N\frac1{r^2}<&\,\int_2^N x^{-2}\,\mathrm{d}x\\ =&\,\frac12-\frac1N. \end{align*}

Adding 1+141+\frac14 gives

∑r=1N1r2<74−1N.\sum_{r=1}^N\frac1{r^2}<\boxed{\frac74-\frac1N}.

(c)

解法一

思路

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部分和递增且有上界,因此收敛。让 NN 趋于无穷,首尾的 1/N1/N 项消失。若要保留严格不等号,可保留 [2,3][2,3] 上固定的矩形误差。

答题过程

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The partial sums increase and are bounded above by 74\frac74, so they converge. In (a) the first rectangle exceeds the integral on [2,3][2,3] by 14−16=112\frac14-\frac16=\frac1{12}. In (b) the integral on that same interval exceeds the first rectangle by 16−19=118\frac16-\frac19=\frac1{18}. These positive gaps remain as N→∞N\to\infty. In particular,

32<∑r=1∞1r2<74.\boxed{\frac32<\sum_{r=1}^{\infty}\frac1{r^2}<\frac74}.