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P2 2025 May A Q5

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 5

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

The sequence u1,u2,u3,…u_1,u_2,u_3,\ldots is defined by

un+1=1−1unu_{n+1}=1-\frac1{u_n} u1=4u_1=4

(a) Show that this is a periodic sequence of order 33

(3)

(b) Find the value of

∑n=1180(5n+3+un)\sum_{n=1}^{180}(5n+3+u_n)
(4)
题目中文翻译

本题必须展示计算过程的所有步骤。

依赖计算器技术的解答不予接受。

数列 u1,u2,u3,…u_1,u_2,u_3,\ldots 定义为

un+1=1−1unu_{n+1}=1-\frac1{u_n}

且

u1=4u_1=4

(a) 证明这是一个周期为 33 的数列。

(3)

(b) 求

∑n=1180(5n+3+un)\sum_{n=1}^{180}(5n+3+u_n)

的值。

(4)

解答

(a)

解法一

思路

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从 u1=4u_1=4 开始连续代入递推式。如果 u4=u1u_4=u_1，而前三项没有更早重复，就说明周期是 33。

答题过程

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Given

u1=4,\begin{align*} u_1=4, \end{align*}

we have

u2= 1−14=34,u3= 1−43=−13,u4= 1−(−3)=4.\begin{align*} u_2=&\,1-\frac14=\frac34,\\ u_3=&\,1-\frac43=-\frac13,\\ u_4=&\,1-(-3)=4. \end{align*}

Since u4=u1u_4=u_1 and the same recurrence is used each time, the sequence repeats. The first three terms are distinct, so the period is exactly 33:

u1= 4,u2=34,u3=−13,u4= 4,u5=34,u6=−13,…\begin{align*} u_1=&\,4,\quad u_2=\frac34,\quad u_3=-\frac13,\\ u_4=&\,4,\quad u_5=\frac34,\quad u_6=-\frac13,\quad\ldots \end{align*}

Therefore, this is a periodic sequence of order 33.

(b)

解法一

思路

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把求和拆成两部分：

∑(5n+3)\sum(5n+3) 是等差数列求和；∑un\sum u_n 每 33 项一组重复，180180 项正好是 6060 组。

答题过程

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First,

∑n=1180(5n+3)= 1802(8+903)= 90(911)= 81990.\begin{align*} \sum_{n=1}^{180}(5n+3) =&\,\frac{180}{2}(8+903)\\ =&\,90(911)\\ =&\,81990. \end{align*}

One cycle of unu_n has sum

4+34−13= 48+9−412= 5312.\begin{align*} 4+\frac34-\frac13=&\,\frac{48+9-4}{12}\\ =&\,\frac{53}{12}. \end{align*}

Since

180=60×3,\begin{align*} 180=60\times3, \end{align*}

we get

∑n=1180un= 60(5312)= 265.\begin{align*} \sum_{n=1}^{180}u_n =&\,60\bigg(\frac{53}{12}\bigg)\\ =&\,265. \end{align*}

Therefore,

 ∑n=1180(5n+3+un)= 81990+265= 82255.\begin{align*} &\,\sum_{n=1}^{180}(5n+3+u_n)\\ =&\,81990+265\\ =&\,82255. \end{align*}

解法二

思路

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使用 ∑n=1Nn=N(N+1)2\sum_{n=1}^N n=\frac{N(N+1)}2 计算线性部分，再加上 6060 个周期的数列项之和。这样不需要另求等差数列的首项和末项。

答题过程

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Using the formula for the sum of the first 180180 positive integers,

∑n=1180n=180⋅1812=16290.\sum_{n=1}^{180}n=\frac{180\cdot181}{2}=16290.

Hence

 ∑n=1180(5n+3)= 5∑n=1180n+3(180)= 5(16290)+540= 81990.\begin{align*} &\,\sum_{n=1}^{180}(5n+3)\\ =&\,5\sum_{n=1}^{180}n+3(180)\\ =&\,5(16290)+540\\ =&\,81990. \end{align*}

From part (a), there are 6060 cycles of length 33. One cycle has sum 5312\frac{53}{12}, so

∑n=1180un=60⋅5312=265.\sum_{n=1}^{180}u_n=60\cdot\frac{53}{12}=265.

Therefore,

 ∑n=1180(5n+3+un)= 81990+265= 82255.\begin{align*} &\,\sum_{n=1}^{180}(5n+3+u_n)\\ =&\,81990+265\\ =&\,82255. \end{align*}