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CIE 9231 2024 November Paper 12 Q1

A Level / CIE / FP1

CIE 9231 2024 Nov Paper 12 Paper · Question 1

题目

Problem

The sequence u1u_1, u2u_2, u3u_3, …\ldots is such that u1=4u_1=4 and un+1=3un−2u_{n+1}=3u_n-2 for n⩾1n\geqslant1.

Prove by induction that un=3n+1u_n=3^n+1 for all positive integers nn.

(5)
题目中文翻译

数列 u1u_1、u2u_2、u3u_3、…\ldots 满足 u1=4u_1=4,且当 n⩾1n\geqslant1 时,un+1=3un−2u_{n+1}=3u_n-2。

用数学归纳法证明:对所有正整数 nn,un=3n+1u_n=3^n+1。

(5)
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

解法一

思路

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验证首项,再假设第k项成立并代入递推式,推出第k+1项的目标形式。

答题过程

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For n=1n=1, 31+1=4=u13^1+1=4=u_1, so the formula is true.

Assume that uk=3k+1u_k=3^k+1 for some positive integer kk. Using the recurrence,

uk+1= 3uk−2= 3(3k+1)−2= 3k+1+1.\begin{align*} u_{k+1}=&\,3u_k-2\\[4mm] =&\,3(3^k+1)-2\\[4mm] =&\,3^{k+1}+1. \end{align*}

Hence truth for n=kn=k implies truth for n=k+1n=k+1. Since the formula holds for n=1n=1, by mathematical induction

un=3n+1\boxed{u_n=3^n+1}

for every positive integer nn.