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FP1 2021 Oct Q9

A Level / Edexcel / FP1

IAL 2021 Oct Paper · Question 9

题目

Problem

9. (i) A sequence of numbers is defined by

u1= 0,u2= −6\begin{align*} u_1=&\,0,\\ u_2=&\,-6 \end{align*}  un+2= 5un+1−6un, n≥1\begin{align*} &\,u_{n+2}\\ =&\,5u_{n+1}-6u_n,\\ &\,n\geq1 \end{align*}

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

un= 3×2n −2×3n\begin{align*} u_n=&\,3\times2^n\\ &\,-2\times3^n \end{align*}
(5)

(ii) Prove by induction that, for all positive integers nn,

f(n)= 33n−2 +24n−1\begin{align*} f(n)=&\,3^{3n-2}\\ &\,+2^{4n-1} \end{align*}

is divisible by 1111

(5)
题目中文翻译
  1. (i) 一个数列定义如下：
u1= 0,u2= −6\begin{align*} u_1=&\,0,\\ u_2=&\,-6 \end{align*}  un+2= 5un+1−6un, n≥1\begin{align*} &\,u_{n+2}\\ =&\,5u_{n+1}-6u_n,\\ &\,n\geq1 \end{align*}

用数学归纳法证明，对于 n∈Z+n \in \mathbb{Z}^+

un= 3×2n −2×3n\begin{align*} u_n=&\,3\times2^n\\ &\,-2\times3^n \end{align*}
(5)

(ii) 用数学归纳法证明，对于所有正整数 nn

f(n)= 33n−2 +24n−1\begin{align*} f(n)=&\,3^{3n-2}\\ &\,+2^{4n-1} \end{align*}

能被 1111 整除。

(5)

解答

(i)

解法一

思路

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递推关系同时依赖前两项，因此归纳基础必须验证 n=1n=1 和 n=2n=2。归纳步骤中假设公式对 n=kn=k、n=k+1n=k+1 都成立，将两式代入递推关系，证明公式对 n=k+2n=k+2 成立。

答题过程

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

 3×21−2×31= 6−6= 0= u1.\begin{align*} &\,3\times2^1-2\times3^1\\ =&\,6-6\\ =&\,0\\ =&\,u_1. \end{align*}

For n=2n=2, it gives

 3×22−2×32= 12−18= −6= u2.\begin{align*} &\,3\times2^2-2\times3^2\\ =&\,12-18\\ =&\,-6\\ =&\,u_2. \end{align*}

Thus 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. Then

uk= 3×2k −2×3k,uk+1= 3×2k+1 −2×3k+1.\begin{align*} u_k=&\,3\times2^k\\ &\,-2\times3^k,\\ u_{k+1}=&\,3\times2^{k+1}\\ &\,-2\times3^{k+1}. \end{align*}

Using the recurrence relation,

 uk+2= 5uk+1−6uk= 5(3×2k+1−2×3k+1) −6(3×2k−2×3k)= 15×2k+1−10×3k+1 −18×2k+12×3k= 30×2k−30×3k −18×2k+12×3k= (30−18)×2k +(−30+12)×3k= 12×2k−18×3k= 3×4×2k −2×9×3k= 3×2k+2−2×3k+2.\begin{align*} &\,u_{k+2}\\ =&\,5u_{k+1}-6u_k\\ =&\,5\big(3\times2^{k+1}-2\times3^{k+1}\big)\\ &\,\hspace{2pt}-6\big(3\times2^k-2\times3^k\big)\\ =&\,15\times2^{k+1}-10\times3^{k+1}\\ &\,\hspace{2pt}-18\times2^k+12\times3^k\\ =&\,30\times2^k-30\times3^k\\ &\,\hspace{2pt}-18\times2^k+12\times3^k\\ =&\,(30-18)\times2^k\\ &\,\hspace{2pt}+(-30+12)\times3^k\\ =&\,12\times2^k-18\times3^k\\ =&\,3\times4\times2^k\\ &\,\hspace{2pt}-2\times9\times3^k\\ =&\,3\times2^{k+2}-2\times3^{k+2}. \end{align*}

Therefore, if the result is true for n=kn=k and n=k+1n=k+1, it is true for n=k+2n=k+2. Since it is true for n=1n=1 and n=2n=2, the result is true for all n∈Z+n\in\mathbb{Z}^+ by mathematical induction.

(ii)

解法一

思路

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先验证 n=1n=1。假设 f(k)f(k) 可被 1111 整除，把 f(k+1)f(k+1) 中的指数降到与 f(k)f(k) 一致，再将系数 2727 拆成 16+1116+11，配出 16f(k)16f(k)。

答题过程

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For n=1n=1,

f(1)= 31+23= 3+8= 11,\begin{align*} f(1)=&\,3^1+2^3\\ =&\,3+8\\ =&\,11, \end{align*}

which is divisible by 1111.

Assume that the result is true for some positive integer kk, so

f(k)=33k−2+24k−1f(k)=3^{3k-2}+2^{4k-1}

is divisible by 1111.

Now

 f(k+1)= 33k+1+24k+3= 27×33k−2 +16×24k−1= (16+11)×33k−2 +16×24k−1= 16(33k−2+24k−1) +11×33k−2= 16f(k)+11×33k−2.\begin{align*} &\,f(k+1)\\ =&\,3^{3k+1}+2^{4k+3}\\ =&\,27\times3^{3k-2}\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,(16+11)\times3^{3k-2}\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,16\big(3^{3k-2}+2^{4k-1}\big)\\ &\,\hspace{2pt}+11\times3^{3k-2}\\ =&\,16f(k)+11\times3^{3k-2}. \end{align*}

By the induction hypothesis, 16f(k)16f(k) is divisible by 1111, and 11×33k−211\times3^{3k-2} is also divisible by 1111. Therefore, f(k+1)f(k+1) is divisible by 1111.

Hence, if the result is true for n=kn=k, it is true for n=k+1n=k+1. Since it is true for n=1n=1, f(n)f(n) is divisible by 1111 for all positive integers nn, by mathematical induction.

解法二

思路

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把系数 1616 拆成 27−1127-11，配出 27f(k)27f(k)。这样 f(k+1)f(k+1) 就是两个可被 1111 整除的数之差。

答题过程

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For n=1n=1,

f(1)= 31+23= 3+8= 11,\begin{align*} f(1)=&\,3^1+2^3\\ =&\,3+8\\ =&\,11, \end{align*}

which is divisible by 1111.

Assume that the result is true for some positive integer kk, so

f(k)=33k−2+24k−1f(k)=3^{3k-2}+2^{4k-1}

is divisible by 1111.

Then

 f(k+1)= 33k+1+24k+3= 27×33k−2 +16×24k−1= 27×33k−2 +(27−11)×24k−1= 27(33k−2+24k−1) −11×24k−1= 27f(k)−11×24k−1.\begin{align*} &\,f(k+1)\\ =&\,3^{3k+1}+2^{4k+3}\\ =&\,27\times3^{3k-2}\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,27\times3^{3k-2}\\ &\,\hspace{2pt}+(27-11)\times2^{4k-1}\\ =&\,27\big(3^{3k-2}+2^{4k-1}\big)\\ &\,\hspace{2pt}-11\times2^{4k-1}\\ =&\,27f(k)-11\times2^{4k-1}. \end{align*}

By the induction hypothesis, 27f(k)27f(k) is divisible by 1111. The term 11×24k−111\times2^{4k-1} is also divisible by 1111, so f(k+1)f(k+1) is divisible by 1111.

Thus, if the result is true for n=kn=k, it is true for n=k+1n=k+1. Since it is true for n=1n=1, the result follows for all positive integers nn by mathematical induction.

解法三

思路

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把归纳假设写成 f(k)=11Mf(k)=11M，其中 MM 是整数。从中表示出 33k−23^{3k-2}，代入 f(k+1)f(k+1) 后直接提出因子 1111。

答题过程

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For n=1n=1,

f(1)= 31+23= 3+8= 11,\begin{align*} f(1)=&\,3^1+2^3\\ =&\,3+8\\ =&\,11, \end{align*}

so the result is true.

Assume that the result is true for some positive integer kk. Then there is an integer MM such that

f(k)=11M.f(k)=11M.

Thus

33k−2+24k−1=11M,3^{3k-2}+2^{4k-1}=11M,

so

33k−2=11M−24k−1.3^{3k-2}=11M-2^{4k-1}.

Substituting into f(k+1)f(k+1),

 f(k+1)= 33k+1+24k+3= 27×33k−2 +16×24k−1= 27(11M−24k−1) +16×24k−1= 297M−27×24k−1 +16×24k−1= 297M +(−27+16)×24k−1= 297M−11×24k−1= 11(27M−24k−1).\begin{align*} &\,f(k+1)\\ =&\,3^{3k+1}+2^{4k+3}\\ =&\,27\times3^{3k-2}\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,27\big(11M-2^{4k-1}\big)\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,297M-27\times2^{4k-1}\\ &\,\hspace{2pt}+16\times2^{4k-1}\\ =&\,297M\\ &\,\hspace{2pt}+(-27+16)\times2^{4k-1}\\ =&\,297M-11\times2^{4k-1}\\ =&\,11\big(27M-2^{4k-1}\big). \end{align*}

Since MM and 24k−12^{4k-1} are integers, 27M−24k−127M-2^{4k-1} is an integer. Hence f(k+1)f(k+1) is divisible by 1111.

Therefore, if the result is true for n=kn=k, it is true for n=k+1n=k+1. Since it is true for n=1n=1, f(n)f(n) is divisible by 1111 for all positive integers nn, by mathematical induction.