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FP1 2026 Jan Q10

A Level / Edexcel / FP1

IAL 2026 Jan Paper · Question 10

题目

Problem

A sequence of numbers u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1= 8u2= 10\begin{align*} u_1=&\,8\\ u_2=&\,10 \end{align*} un+2=un+1+2unu_{n+2}=u_{n+1}+2u_n

n∈Z+n\in\mathbb{Z}^+

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

 un= 3×2n−2×(−1)n\begin{align*} &\,u_n\\ =&\,3\times2^n-2\times(-1)^n \end{align*}
(6)
题目中文翻译

一个数列 u1,u2,u3,…u_1, u_2, u_3, \ldots 由下式定义

u1= 8u2= 10\begin{align*} u_1=&\,8\\ u_2=&\,10 \end{align*} un+2=un+1+2unu_{n+2}=u_{n+1}+2u_n

n∈Z+n\in\mathbb{Z}^+

用数学归纳法证明,对于 n∈Z+n \in \mathbb{Z}^+

 un= 3×2n−2×(−1)n\begin{align*} &\,u_n\\ =&\,3\times2^n-2\times(-1)^n \end{align*}
(6)

解答

解法一

思路

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递推式同时使用前两项,所以归纳证明必须先验证 n=1n=1 与 n=2n=2 两个基例。归纳步骤中假设公式对相邻的 uku_k、uk+1u_{k+1} 成立,再代入递推式推出 uk+2u_{k+2} 的目标形式。

答题过程

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For n=1n=1, the proposed formula gives

 3×21−2×(−1)1= 6+2= 8=u1.\begin{align*} &\,3\times2^1-2\times(-1)^1\\ =&\,6+2\\ =&\,8=u_1. \end{align*}

For n=2n=2, it gives

 3×22−2×(−1)2= 12−2= 10=u2.\begin{align*} &\,3\times2^2-2\times(-1)^2\\ =&\,12-2\\ =&\,10=u_2. \end{align*}

Therefore, the result is true for n=1n=1 and n=2n=2.

Assume that the result is true for two consecutive positive integers kk and k+1k+1. Thus,

uk=3×2k−2×(−1)ku_k=3\times2^k-2\times(-1)^k

and

 uk+1= 3×2k+1 −2×(−1)k+1.\begin{align*} &\,u_{k+1}\\ =&\,3\times2^{k+1}\\ &\,\hspace{2pt}-2\times(-1)^{k+1}. \end{align*}

Using the recurrence relation,

 uk+2= uk+1+2uk= 3×2k+1 −2×(−1)k+1 +2[3×2k −2×(−1)k]= 3×2k+1+6×2k −2×(−1)k+1 −4×(−1)k.\begin{align*} &\,u_{k+2}\\ =&\,u_{k+1}+2u_k\\ =&\,3\times2^{k+1}\\ &\,\hspace{2pt}-2\times(-1)^{k+1}\\ &\,\hspace{4pt}+2\big[3\times2^k\\ &\,\hspace{6pt}-2\times(-1)^k\big]\\ =&\,3\times2^{k+1}+6\times2^k\\ &\,\hspace{2pt}-2\times(-1)^{k+1}\\ &\,\hspace{4pt}-4\times(-1)^k. \end{align*}

The powers of 22 combine as

 3×2k+1+6×2k= (6+6)2k= 12×2k= 3×2k+2.\begin{align*} &\,3\times2^{k+1}+6\times2^k\\ =&\,(6+6)2^k\\ =&\,12\times2^k\\ =&\,3\times2^{k+2}. \end{align*}

Also,

 −2×(−1)k+1 −4×(−1)k= 2×(−1)k −4×(−1)k= −2×(−1)k.\begin{align*} &\,-2\times(-1)^{k+1}\\ &\,\hspace{2pt}-4\times(-1)^k\\ =&\,2\times(-1)^k\\ &\,\hspace{2pt}-4\times(-1)^k\\ =&\,-2\times(-1)^k. \end{align*}

Since (−1)k+2=(−1)k(-1)^{k+2}=(-1)^k, it follows that

 uk+2= 3×2k+2−2×(−1)k= 3×2k+2 −2×(−1)k+2.\begin{align*} &\,u_{k+2}\\ =&\,3\times2^{k+2}-2\times(-1)^k\\ =&\,3\times2^{k+2}\\ &\,\hspace{2pt}-2\times(-1)^{k+2}. \end{align*}

Hence, 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 induction,

un= 3×2n−2×(−1)n\begin{align*} u_n=&\,3\times2^n-2\times(-1)^n \end{align*}

for all n∈Z+n\in\mathbb{Z}^+.