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CIE 9231 2025 November Paper 14 Q1

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 14 Paper · Question 1

题目

Problem

Prove by mathematical induction that (65)n⩾1+15n\big(\frac65\big)^n\geqslant1+\frac15n for all positive integers nn.

(5)

题目中文翻译

利用数学归纳法证明:对所有正整数 nn,都有 (65)n⩾1+15n\big(\frac65\big)^n\geqslant1+\frac15n。

(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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先检查n=1等号成立。归纳步骤乘正数6/5后,分离出目标下界与非负余项。

答题过程

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For n=1n=1, both sides equal 6/56/5, so the statement is true.

Assume that (6/5)k≥1+k/5(6/5)^k\ge1+k/5 for some positive integer kk. Multiplying by 6/5>06/5>0,

(65)k+1≥ 65(1+k5)= 1+k+15+k25≥ 1+k+15.\begin{align*} \bigg(\frac65\bigg)^{k+1} \ge&\,\frac65\bigg(1+\frac k5\bigg) \\[4mm] =&\,1+\frac{k+1}{5}+\frac k{25} \\[4mm] \ge&\,1+\frac{k+1}{5}. \end{align*}

Thus the statement holds for k+1k+1. By mathematical induction,

(65)n≥1+n5\boxed{\bigg(\frac65\bigg)^n\ge1+\frac n5}

for every positive integer nn.