题目
Problem
Prove by mathematical induction that, for all positive integers n,
dxndn(x2ex)=(x2+2nx+n(n−1))ex.
(6)
题目中文翻译
用数学归纳法证明,对于所有正整数 n,
dxndn(x2ex)=(x2+2nx+n(n−1))ex.
(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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用乘积法则验证第一阶导数,再在归纳步骤把指数函数提出,合并多项式系数。
答题过程
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For n=1,
dxd(x2ex)=(x2+2x)ex,
which agrees with the given formula, so the result is true.
Assume that for some positive integer k,
dxkdk(x2ex)=(x2+2kx+k(k−1))ex.
Differentiating using the product rule,
==dxk+1dk+1(x2ex)[x2+2kx+k(k−1)+2x+2k]ex[x2+2(k+1)x+k(k+1)]ex.
This is the required expression for n=k+1. Since it holds for n=1 and truth for n=k implies truth for n=k+1, the statement follows by mathematical induction for all positive integers n.