CalcGospel 國際數學圖譜
返回

P2 2025 May Q7

A Level / Edexcel / P2

IAL 2025 May Paper · Question 7

题目

Problem

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

(i) A geometric series begins

10+8+6.4+⋯10+8+6.4+\cdots

(a) Find the sum to infinity of this series.

(2)

Given that the kkth term of this series is less than 0.00050.0005

(b) use algebra to find the smallest possible value of kk.

(3)

(ii) An arithmetic series begins

850+843+836+⋯850+843+836+\cdots

Given that the sum of the first nn terms of this series is SnS_n

find the greatest possible value of SnS_n

(4)
题目中文翻译

本题必须展示所有解题步骤。完全依赖计算器技术的解答不予接受。

(i) 一个等比级数的开头为 10+8+6.4+⋯10+8+6.4+\cdots。

(a) 求这个级数的无穷和。（2分）

已知这个级数的第 kk 项小于 0.00050.0005，

(b) 用代数方法求 kk 的最小可能值。（3分）

(ii) 一个等差级数的开头为 850+843+836+⋯850+843+836+\cdots。已知这个级数的前 nn 项之和为 SnS_n，求 SnS_n 的最大可能值。（4分）

解答

(i)(a)

解法一

思路

展开

这是等比级数，首项 a=10a=10，公比 r=810=0.8r=\frac{8}{10}=0.8。因为 ∣r∣<1|r|<1，可以用 sum to infinity 公式。

答题过程

展开

The first term is

a=10,\begin{align*} a=10, \end{align*}

and the common ratio is

r=810=0.8.\begin{align*} r=\frac{8}{10}=0.8. \end{align*}

Since ∣r∣<1|r|<1,

S∞= a1−r= 101−0.8= 50.\begin{align*} S_\infty=&\,\frac{a}{1-r}\\ =&\,\frac{10}{1-0.8}\\ =&\,50. \end{align*}

(i)(b)

解法一

思路

展开

第 kk 项是 10(0.8)k−110(0.8)^{k-1}。题目要它小于 0.00050.0005，用对数解指数不等式。因为 log⁡(0.8)<0\log(0.8)<0，除以它时不等号方向要改变。

答题过程

展开

The kkth term is

10(0.8)k−1.\begin{align*} 10(0.8)^{k-1}. \end{align*}

We need

10(0.8)k−1<0.0005(0.8)k−1<0.00005.\begin{align*} 10(0.8)^{k-1}&<0.0005\\ (0.8)^{k-1}&<0.00005. \end{align*}

Taking logarithms,

(k−1)log⁡(0.8)<log⁡(0.00005).\begin{align*} (k-1)\log(0.8)<\log(0.00005). \end{align*}

Since log⁡(0.8)<0\log(0.8)<0,

k−1>log⁡(0.00005)log⁡(0.8)k−1>44.38….\begin{align*} k-1&>\frac{\log(0.00005)}{\log(0.8)}\\ k-1&>44.38\ldots. \end{align*}

So

k>45.38….\begin{align*} k>45.38\ldots. \end{align*}

The smallest possible integer value is

k=46.\begin{align*} k=46. \end{align*}

(ii)

解法一

思路

展开

这个 arithmetic series 的项先为正，后来变负。要让和最大，就应该加到最后一个正项为止。

答题过程

展开

The arithmetic series has

a=850,d=−7.\begin{align*} a=850,\qquad d=-7. \end{align*}

The nnth term is

un=850+(n−1)(−7).\begin{align*} u_n=850+(n-1)(-7). \end{align*}

Find where the terms change from positive to negative:

850−7(n−1)= 0857−7n= 0n= 8577= 122.428….\begin{align*} 850-7(n-1)=&\,0\\ 857-7n=&\,0\\ n=&\,\frac{857}{7}\\ =&\,122.428\ldots. \end{align*}

In particular, u122=850−7(121)=3>0u_{122}=850-7(121)=3>0 and u123=850−7(122)=−4<0u_{123}=850-7(122)=-4<0.

So the greatest sum is obtained when

n=122.\begin{align*} n=122. \end{align*}

Now

S122= 1222{2(850)+(122−1)(−7)}= 61(1700−847)= 61(853)= 52033.\begin{align*} S_{122} =&\,\frac{122}{2}\{2(850)+(122-1)(-7)\}\\ =&\,61(1700-847)\\ =&\,61(853)\\ =&\,52033. \end{align*}

Therefore, the greatest possible value of SnS_n is

52033.\begin{align*} 52033. \end{align*}

解法二

思路

展开

也可以把 SnS_n 写成关于 nn 的二次式。它是一条开口向下的抛物线，最大值在顶点附近；因为 nn 必须是整数，再选最合适的整数。

答题过程

展开

Using the sum formula,

Sn= n2{2(850)+(n−1)(−7)}= n2(1707−7n)= 17072n−72n2.\begin{align*} S_n =&\,\frac{n}{2}\{2(850)+(n-1)(-7)\}\\ =&\,\frac{n}{2}(1707-7n)\\ =&\,\frac{1707}{2}n-\frac72n^2. \end{align*}

This quadratic has a maximum at

n= −b2a= 1707/27= 170714= 121.928….\begin{align*} n=&\,-\frac{b}{2a}\\ =&\,\frac{1707/2}{7}\\ =&\,\frac{1707}{14}\\ =&\,121.928\ldots. \end{align*}

Since the quadratic opens downwards, its value decreases with distance from the vertex. The nearest integer to 170714\frac{1707}{14} is 122122.

For a direct check, S121=52030S_{121}=52030, S122=52033S_{122}=52033 and S123=52029S_{123}=52029. Hence use n=122n=122.

Then

S122= 1222{2(850)+121(−7)}= 52033.\begin{align*} S_{122} =&\,\frac{122}{2}\{2(850)+121(-7)\}\\ =&\,52033. \end{align*}

Therefore, the greatest possible value of SnS_n is 5203352033.

解法三

思路

展开

把求和公式暂时视为实变量的二次函数，用微分找到最大值的位置，再回到正整数项数比较相邻整数。

答题过程

展开

Write

S(n)= n2[1700−7(n−1)]= 17072n−72n2.\begin{align*} S(n)=&\,\frac n2\big[1700-7(n-1)\big]\\ =&\,\frac{1707}{2}n-\frac72n^2. \end{align*}

Treat nn temporarily as a real variable. Then

dSdn= 17072−7n,d2Sdn2= −7<0.\begin{align*} \frac{\mathrm{d}S}{\mathrm{d}n}=&\,\frac{1707}{2}-7n,\\ \frac{\mathrm{d}^2S}{\mathrm{d}n^2}=&\,-7<0. \end{align*}

Setting the first derivative to zero gives

7n= 17072n= 170714≈ 121.9286.\begin{align*} 7n=&\,\frac{1707}{2}\\ n=&\,\frac{1707}{14}\\ \approx&\,121.9286. \end{align*}

The real-variable quadratic increases up to this point and decreases afterwards. For positive integer nn, the maximum must therefore be at 121121 or 122122.

S121= 1212(1707−7⋅121)= 52030,S122= 1222(1707−7⋅122)= 52033.\begin{align*} S_{121}=&\,\frac{121}{2}(1707-7\cdot121)\\ =&\,52030,\\ S_{122}=&\,\frac{122}{2}(1707-7\cdot122)\\ =&\,52033. \end{align*}

Hence the greatest possible sum is 5203352033.

求顶点的另外两种同骨架检查：配方可得

S(n)= −72(n−170714)2 +1707256.\begin{align*} S(n)=&\,-\frac72\bigg(n-\frac{1707}{14}\bigg)^2\\ &\,\hspace{2pt}+\frac{1707^2}{56}. \end{align*}

或由两个零点 00、17077\frac{1707}{7} 的中点得到同一顶点位置，不必重复求和过程。