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CIE 9231 2023 November Paper 11 Q2

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 11 Paper · Question 2

题目

Problem

Prove by mathematical induction that, for all positive integers nn,

1+2x+3x2+…+nxn−1=1−(n+1)xn+nxn+1(1−x)2.1+2x+3x^2+\ldots+nx^{n-1}=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}.
(6)
题目中文翻译

用数学归纳法证明：对所有正整数 nn，

1+2x+3x2+…+nxn−1=1−(n+1)xn+nxn+1(1−x)2.1+2x+3x^2+\ldots+nx^{n-1}=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}.
(6)
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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固定x不等于1，先验证首项。归纳时加上第k+1项并通分，明确合并相同幂次的系数。

答题过程

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Let x≠1x\ne1. For n=1n=1, the left-hand side is 11 and the right-hand side is

1−2x+x2(1−x)2=1,\frac{1-2x+x^2}{(1-x)^2}=1,

so the identity holds.

Assume that for some positive integer kk,

∑r=1krxr−1=1−(k+1)xk+kxk+1(1−x)2.\sum_{r=1}^k rx^{r-1} =\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}.

Then

∑r=1k+1rxr−1= 1−(k+1)xk+kxk+1(1−x)2 +(k+1)xk= 1−(k+1)xk+kxk+1(1−x)2 +(k+1)xk(1−2x+x2)(1−x)2= 1+[k−2(k+1)]xk+1+(k+1)xk+2(1−x)2= 1−(k+2)xk+1+(k+1)xk+2(1−x)2.\begin{align*} \sum_{r=1}^{k+1}rx^{r-1} =&\,\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}\\[4mm] &\,+(k+1)x^k\\[4mm] =&\,\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}\\[4mm] &\,\hspace{2pt}+\frac{(k+1)x^k(1-2x+x^2)}{(1-x)^2}\\[4mm] =&\,\frac{1+[k-2(k+1)]x^{k+1}+(k+1)x^{k+2}}{(1-x)^2}\\[4mm] =&\,\frac{1-(k+2)x^{k+1}+(k+1)x^{k+2}}{(1-x)^2}. \end{align*}

This is the required expression with n=k+1n=k+1. Since the identity holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, it holds for every positive integer nn, for x≠1x\ne1 as required for the displayed fraction to be defined.