题目
Problem
(i) Prove by induction that, for n∈Z+
=(−2−194)n(1−3n−n9n3n+1)
(5)
(ii) A sequence of numbers is defined by
u1=1,u2=4,un+2=6un+1−9un,n≥1
Prove by induction that, for n∈Z+
un=3n−2(n+2)
(5)
题目中文翻译
(i) 用数学归纳法证明,对于所有正整数 n∈Z+
=(−2−194)n(1−3n−n9n3n+1)
(5)
(ii) 一个数列定义如下:
u1=1,u2=4,un+2=6un+1−9un,n≥1
用数学归纳法证明,对于所有正整数 n∈Z+,
un=3n−2(n+2)
(5)
解答
(i)
解法一
思路
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先验证 n=1。再假设公式对 n=k 成立,把假设中的矩阵右乘原矩阵,逐项计算并整理成把 k 换成 k+1 后的目标矩阵,最后写出完整归纳结论。
答题过程
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Let
A=(−2−194).
For n=1, the right-hand side is
==(1−3−193+1)(−2−194)A.
Hence the result is true for n=1.
Assume that the result is true for n=k, so that
Ak=(1−3k−k9k3k+1).
Then
===Ak+1AkA(1−3k−k9k3k+1)×(−2−194)(−2−3k−k−19+9k3k+4).
The four entries in the product are obtained from
========(1−3k)(−2)+(9k)(−1)−2+6k−9k−2−3k,(1−3k)(9)+(9k)(4)9−27k+36k9+9k,(−k)(−2)+(3k+1)(−1)2k−3k−1−k−1,(−k)(9)+(3k+1)(4)−9k+12k+43k+4.
Each entry has the required form for n=k+1, since
−2−3k=1−3(k+1),9+9k=9(k+1),−k−1=−(k+1),3k+4=3(k+1)+1.
Therefore, if the result is true for n=k, it is also true for n=k+1. Since it is true for n=1, by mathematical induction,
=(−2−194)n(1−3n−n9n3n+1)
for all n∈Z+.
(ii)
解法一
思路
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这是二阶递推数列,归纳时必须验证 n=1,2 两个初值,并同时假设公式对相邻的 k、k+1 成立。把这两个假设代入递推式后,整理成公式在 n=k+2 时的形式。
答题过程
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For n=1,
==31−2(1+2)3−1⋅31=u1.
For n=2,
==32−2(2+2)30⋅44=u2.
Hence the result is true for n=1 and n=2.
Assume that the result is true for n=k and n=k+1, so that
uk=3k−2(k+2)
and
uk+1=3k−1(k+3).
For the powers of 3, note that
==6×3k−12×3×3k−12×3k,
and
==9×3k−232×3k−23k.
Using the recurrence relation,
=====uk+26uk+1−9uk6[3k−1(k+3)]−9[3k−2(k+2)]3k[2(k+3)−(k+2)]3k(k+4)3(k+2)−2×((k+2)+2).
Therefore, if the result is true for n=k and n=k+1, it is also true for n=k+2. Since it is true for n=1 and n=2, by mathematical induction,
un=3n−2(n+2)
for all n∈Z+.