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IAL 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)

解答

(i)(a)

解法一

思路

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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

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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

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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

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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

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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

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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*}

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

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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

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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= −170722(−72)= 170714= 121.928….\begin{align*} n=&\,-\frac{b}{2a}\\ =&\,-\frac{\frac{1707}{2}}{2\left(-\frac72\right)}\\ =&\,\frac{1707}{14}\\ =&\,121.928\ldots. \end{align*}

Since nn must be an integer, 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.