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FP1 2025 Jan Q8

A Level / Edexcel / FP1

IAL 2025 Jan Paper · Question 8

题目

Problem

(i) Prove by induction that, for n∈Z+n \in \mathbb{Z}^+

 (−29−14)n= (1−3n9n−n3n+1)\begin{align*} &\,\begin{pmatrix}-2&9\\-1&4\end{pmatrix}^{n}\\ =&\,\begin{pmatrix}1-3n&9n\\-n&3n+1\end{pmatrix} \end{align*}
(5)

(ii) A sequence of numbers is defined by

u1=1,u2=4,un+2=6un+1−9un,n≥1\begin{gathered} u_1=1,\qquad u_2=4,\\ u_{n+2}=6u_{n+1}-9u_n,\\ n\geq1 \end{gathered}

Prove by induction that, for n∈Z+n \in \mathbb{Z}^+

un=3n−2(n+2)u_n = 3^{n-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+n \in \mathbb{Z}^+

 (−29−14)n= (1−3n9n−n3n+1)\begin{align*} &\,\begin{pmatrix}-2&9\\-1&4\end{pmatrix}^{n}\\ =&\,\begin{pmatrix}1-3n&9n\\-n&3n+1\end{pmatrix} \end{align*}
(5)

(ii) 一个数列定义如下:

u1=1,u2=4,un+2=6un+1−9un,n≥1\begin{gathered} u_1=1,\qquad u_2=4,\\ u_{n+2}=6u_{n+1}-9u_n,\\ n\geq1 \end{gathered}

用数学归纳法证明,对于所有正整数 n∈Z+n \in \mathbb{Z}^+,

un=3n−2(n+2)u_n = 3^{n-2}(n + 2)
(5)

解答

(i)

解法一

思路

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先验证 n=1n=1。再假设公式对 n=kn=k 成立,把假设中的矩阵右乘原矩阵,逐项计算并整理成把 kk 换成 k+1k+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=(−29−14).A=\begin{pmatrix}-2&9\\-1&4\end{pmatrix}.

For n=1n=1, the right-hand side is

 (1−39−13+1)= (−29−14)= A.\begin{align*} &\,\begin{pmatrix} 1-3&9\\ -1&3+1 \end{pmatrix}\\ =&\,\begin{pmatrix} -2&9\\ -1&4 \end{pmatrix}\\ =&\,A. \end{align*}

Hence the result is true for n=1n=1.

Assume that the result is true for n=kn=k, so that

Ak=(1−3k9k−k3k+1).A^k= \begin{pmatrix} 1-3k&9k\\ -k&3k+1 \end{pmatrix}.

Then

 Ak+1= AkA= (1−3k9k−k3k+1) ×(−29−14)= (−2−3k9+9k−k−13k+4).\begin{align*} &\,A^{k+1}\\ =&\,A^kA\\ =&\,\begin{pmatrix} 1-3k&9k\\ -k&3k+1 \end{pmatrix}\\ &\,\hspace{2pt}\times\begin{pmatrix} -2&9\\ -1&4 \end{pmatrix}\\ =&\,\begin{pmatrix} -2-3k&9+9k\\ -k-1&3k+4 \end{pmatrix}. \end{align*}

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+36k= 9+9k, (−k)(−2) +(3k+1)(−1)= 2k−3k−1= −k−1, (−k)(9)+(3k+1)(4)= −9k+12k+4= 3k+4.\begin{align*} &\,(1-3k)(-2)+(9k)(-1)\\ =&\,-2+6k-9k\\ =&\,-2-3k,\\ &\,(1-3k)(9)+(9k)(4)\\ =&\,9-27k+36k\\ =&\,9+9k,\\ &\,(-k)(-2)\\ &\,\hspace{2pt}+(3k+1)(-1)\\ =&\,2k-3k-1\\ =&\,-k-1,\\ &\,(-k)(9)+(3k+1)(4)\\ =&\,-9k+12k+4\\ =&\,3k+4. \end{align*}

Each entry has the required form for n=k+1n=k+1, since

−2−3k=1−3(k+1),9+9k=9(k+1),−k−1=−(k+1),3k+4=3(k+1)+1.\begin{gathered} -2-3k=1-3(k+1),\\ 9+9k=9(k+1),\\ -k-1=-(k+1),\\ 3k+4=3(k+1)+1. \end{gathered}

Therefore, if the result is true for n=kn=k, it is also true for n=k+1n=k+1. Since it is true for n=1n=1, by mathematical induction,

 (−29−14)n= (1−3n9n−n3n+1)\begin{align*} &\,\begin{pmatrix}-2&9\\-1&4\end{pmatrix}^{n}\\ =&\,\begin{pmatrix} 1-3n&9n\\ -n&3n+1 \end{pmatrix} \end{align*}

for all n∈Z+n\in\mathbb 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=1,2n=1,2 两个初值,并同时假设公式对相邻的 kk、k+1k+1 成立。把这两个假设代入递推式后,整理成公式在 n=k+2n=k+2 时的形式。

答题过程

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For n=1n=1,

 31−2(1+2)= 3−1⋅3= 1=u1.\begin{align*} &\,3^{1-2}(1+2)\\ =&\,3^{-1}\cdot3\\ =&\,1=u_1. \end{align*}

For n=2n=2,

 32−2(2+2)= 30⋅4= 4=u2.\begin{align*} &\,3^{2-2}(2+2)\\ =&\,3^0\cdot4\\ =&\,4=u_2. \end{align*}

Hence the result is true for n=1n=1 and n=2n=2.

Assume that the result is true for n=kn=k and n=k+1n=k+1, so that

uk=3k−2(k+2)u_k=3^{k-2}(k+2)

and

uk+1=3k−1(k+3).u_{k+1}=3^{k-1}(k+3).

For the powers of 3, note that

 6×3k−1= 2×3×3k−1= 2×3k,\begin{align*} &\,6\times3^{k-1}\\ =&\,2\times3\times3^{k-1}\\ =&\,2\times3^k, \end{align*}

and

 9×3k−2= 32×3k−2= 3k.\begin{align*} &\,9\times3^{k-2}\\ =&\,3^2\times3^{k-2}\\ =&\,3^k. \end{align*}

Using the recurrence relation,

 uk+2= 6uk+1−9uk= 6[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).\begin{align*} &\,u_{k+2}\\ =&\,6u_{k+1}-9u_k\\ =&\,6\bigl[3^{k-1}(k+3)\bigr]\\ &\,\hspace{2pt}-9\bigl[3^{k-2}(k+2)\bigr]\\ =&\,3^k\bigl[2(k+3)-(k+2)\bigr]\\ =&\,3^k(k+4)\\ =&\,3^{(k+2)-2}\\ &\,\hspace{2pt}\times\big((k+2)+2\big). \end{align*}

Therefore, if the result is true for n=kn=k and n=k+1n=k+1, it is also true for n=k+2n=k+2. Since it is true for n=1n=1 and n=2n=2, by mathematical induction,

un=3n−2(n+2)\boxed{u_n=3^{n-2}(n+2)}

for all n∈Z+n\in\mathbb Z^+.