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FP1 2024 Jan Q10

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 10

题目

Problem

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

 (5−141)n= 3n−1×\begin{align*} &\,\begin{pmatrix}5&-1\\4&1\end{pmatrix}^n\\ =&\,3^{n-1}\times \end{align*} (2n+3−n4n3−2n).\begin{pmatrix}2n+3&-n\\4n&3-2n\end{pmatrix}.
(5)

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

 f(n)= 82n+1+62n−1\begin{align*} &\,f(n)\\ =&\,8^{2n+1}+6^{2n-1} \end{align*}

is divisible by 77

(5)
题目中文翻译

(i) 用数学归纳法证明：对于所有正整数 nn，

 (5−141)n= 3n−1×\begin{align*} &\,\begin{pmatrix}5&-1\\4&1\end{pmatrix}^n\\ =&\,3^{n-1}\times \end{align*} (2n+3−n4n3−2n).\begin{pmatrix}2n+3&-n\\4n&3-2n\end{pmatrix}.
(5)

(ii) 用数学归纳法证明：对于所有正整数 nn，

 f(n)= 82n+1+62n−1\begin{align*} &\,f(n)\\ =&\,8^{2n+1}+6^{2n-1} \end{align*}

能被 77 整除。

(5)

解答

(i)

解法一

思路

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先验证 n=1n=1。归纳步骤中，假设公式对 n=kn=k 成立，再把该矩阵的 kk 次幂右乘原矩阵。矩阵乘法后，每个元素都能提出因数 3，与外面的 3k−13^{k-1} 合并成 3k3^k。

答题过程

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Let

A=(5−141).\mathbf A=\begin{pmatrix}5&-1\\4&1\end{pmatrix}.

For n=1n=1,

2(1)+3= 5,4(1)= 4,3−2(1)= 1.\begin{align*} 2(1)+3=&\,5,\\ 4(1)=&\,4,\\ 3-2(1)=&\,1. \end{align*}

Thus

 31−1 ×(5−141)= (5−141)= A.\begin{align*} &\,3^{1-1}\\ &\,\hspace{2pt}\times\begin{pmatrix} 5&-1\\ 4&1 \end{pmatrix}\\ =&\, \begin{pmatrix}5&-1\\4&1\end{pmatrix}\\ =&\,\mathbf A. \end{align*}

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

Assume that the result is true for n=kn=k, where k∈Z+k\in\mathbb Z^+. Thus

 Ak= 3k−1 ×(2k+3−k4k3−2k).\begin{align*} &\,\mathbf A^k\\ =&\,3^{k-1}\\ &\,\hspace{2pt}\times\begin{pmatrix} 2k+3&-k\\ 4k&3-2k \end{pmatrix}. \end{align*}

For the product of the two matrices, the four entries are

 a= 5(2k+3)+4(−k)= 10k+15−4k= 6k+15,\begin{align*} &\,a\\ =&\,5(2k+3)+4(-k)\\ =&\,10k+15-4k\\ =&\,6k+15, \end{align*}  b= −(2k+3)+(−k)= −2k−3−k= −3k−3,\begin{align*} &\,b\\ =&\,-(2k+3)+(-k)\\ =&\,-2k-3-k\\ =&\,-3k-3, \end{align*}  c= 5(4k)+4(3−2k)= 20k+12−8k= 12k+12,\begin{align*} &\,c\\ =&\,5(4k)+4(3-2k)\\ =&\,20k+12-8k\\ =&\,12k+12, \end{align*}

and

 d= −4k+(3−2k)= 3−6k.\begin{align*} &\,d\\ =&\,-4k+(3-2k)\\ =&\,3-6k. \end{align*}

Let

N=(abcd).\mathbf N=\begin{pmatrix}a&b\\c&d\end{pmatrix}.

Therefore,

 Ak+1= AkA= 3k−1 ×(2k+3−k4k3−2k) ×(5−141)= 3k−1N.\begin{align*} &\,\mathbf A^{k+1}\\ =&\,\mathbf A^k\mathbf A\\ =&\,3^{k-1}\\ &\,\hspace{2pt}\times \begin{pmatrix}2k+3&-k\\4k&3-2k\end{pmatrix}\\ &\,\hspace{4pt}\times \begin{pmatrix}5&-1\\4&1\end{pmatrix}\\ =&\,3^{k-1}\mathbf N. \end{align*}

Each entry of the last matrix has a factor of 3:

6k+15= 3(2k+5),−3k−3= 3(−k−1),12k+12= 3(4k+4),3−6k= 3(1−2k).\begin{align*} 6k+15=&\,3(2k+5),\\ -3k-3=&\,3(-k-1),\\ 12k+12=&\,3(4k+4),\\ 3-6k=&\,3(1-2k). \end{align*}

Since 3k−1×3=3k3^{k-1}\times3=3^k, it follows that

 Ak+1= 3k ×(2k+5−k−14k+41−2k).\begin{align*} &\,\mathbf A^{k+1}\\ =&\,3^k\\ &\,\hspace{2pt}\times \begin{pmatrix}2k+5&-k-1\\4k+4&1-2k\end{pmatrix}. \end{align*}

The entries can be written as

2k+5= 2(k+1)+3,−k−1= −(k+1),4k+4= 4(k+1),1−2k= 3−2(k+1).\begin{align*} 2k+5=&\,2(k+1)+3,\\ -k-1=&\,-(k+1),\\ 4k+4=&\,4(k+1),\\ 1-2k=&\,3-2(k+1). \end{align*}

This is the required formula for n=k+1n=k+1. 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, by mathematical induction,

 An= 3n−1 ×(2n+3−n4n3−2n).\begin{align*} &\,\mathbf A^n\\ =&\,3^{n-1}\\ &\,\hspace{2pt}\times \begin{pmatrix} 2n+3&-n\\ 4n&3-2n \end{pmatrix}. \end{align*}

for all n∈Z+n\in\mathbb Z^+.

(ii)

解法一

思路

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先验证 n=1n=1。归纳步骤中，把 f(k+1)f(k+1) 凑成 64f(k)64f(k) 减去一个含因数 7 的项；利用归纳假设即可证明两部分都能被 7 整除。

答题过程

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For n=1n=1,

f(1)= 83+6= 512+6= 518= 7×74,\begin{align*} f(1)=&\,8^3+6\\ =&\,512+6\\ =&\,518\\ =&\,7\times74, \end{align*}

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

Assume that the result is true for n=kn=k. Thus

f(k)=82k+1+62k−1f(k)=8^{2k+1}+6^{2k-1}

is divisible by 7.

Now,

 f(k+1)= 82k+3+62k+1= 64(82k+1) +36(62k−1)= 64(82k+1) +(64−28)(62k−1)= 64(82k+1+62k−1) −28(62k−1)= 64f(k)−28(62k−1).\begin{align*} &\,f(k+1)\\ =&\,8^{2k+3}+6^{2k+1}\\ =&\,64\big(8^{2k+1}\big)\\ &\,\hspace{2pt}+36\big(6^{2k-1}\big)\\ =&\,64\big(8^{2k+1}\big)\\ &\,\hspace{2pt}+(64-28)\big(6^{2k-1}\big)\\ =&\,64\big(8^{2k+1}+6^{2k-1}\big)\\ &\,\hspace{2pt}-28\big(6^{2k-1}\big)\\ =&\,64f(k)-28\big(6^{2k-1}\big). \end{align*}

By the induction hypothesis, 64f(k)64f(k) is divisible by 7, and 28(62k−1)28\big(6^{2k-1}\big) is divisible by 7. Hence f(k+1)f(k+1) is divisible by 7.

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, by mathematical induction,

7∣f(n) for all n∈Z+.\boxed{7\mid f(n)\text{ for all }n\in\mathbb Z^+}.

解法二

思路

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也可以改为凑出 36f(k)36f(k)。此时剩下的是 28⋅82k+128\cdot8^{2k+1}，同样显然含有因数 7。

答题过程

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For n=1n=1,

f(1)= 83+6= 512+6= 518= 7×74,\begin{align*} f(1)=&\,8^3+6\\ =&\,512+6\\ =&\,518\\ =&\,7\times74, \end{align*}

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

Assume that f(k)=82k+1+62k−1f(k)=8^{2k+1}+6^{2k-1} is divisible by 7. Then

 f(k+1)= 64(82k+1) +36(62k−1)= (36+28)(82k+1) +36(62k−1)= 36(82k+1+62k−1) +28(82k+1)= 36f(k)+28(82k+1).\begin{align*} &\,f(k+1)\\ =&\,64\big(8^{2k+1}\big)\\ &\,\hspace{2pt}+36\big(6^{2k-1}\big)\\ =&\,(36+28)\big(8^{2k+1}\big)\\ &\,\hspace{2pt}+36\big(6^{2k-1}\big)\\ =&\,36\big(8^{2k+1}+6^{2k-1}\big)\\ &\,\hspace{2pt}+28\big(8^{2k+1}\big)\\ =&\,36f(k)+28\big(8^{2k+1}\big). \end{align*}

Both terms are divisible by 7, so f(k+1)f(k+1) is divisible by 7. 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, by mathematical induction,

7∣f(n) for all n∈Z+.\boxed{7\mid f(n)\text{ for all }n\in\mathbb Z^+}.

解法三

思路

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相邻两项作差后，两类幂的系数分别变成 64−1=6364-1=63 与 36−1=3536-1=35，都含有因数 7。再用归纳假设，把“差能被 7 整除”转回到下一项能被 7 整除。

答题过程

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For n=1n=1,

f(1)= 83+6= 518= 7×74.\begin{align*} f(1)=&\,8^3+6\\ =&\,518\\ =&\,7\times74. \end{align*}

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

Assume that the result is true for n=kn=k, where kk is a positive integer. Then f(k)=7mf(k)=7m for some integer mm.

Consider the difference:

 f(k+1)−f(k)= 82k+3+62k+1 −82k+1−62k−1= (64−1)82k+1 +(36−1)62k−1= 63⋅82k+1 +35⋅62k−1.\begin{align*} &\,f(k+1)-f(k)\\ =&\,8^{2k+3}+6^{2k+1}\\ &\,\hspace{2pt}-8^{2k+1}-6^{2k-1}\\ =&\,(64-1)8^{2k+1}\\ &\,\hspace{2pt}+(36-1)6^{2k-1}\\ =&\,63\cdot8^{2k+1}\\ &\,\hspace{2pt}+35\cdot6^{2k-1}. \end{align*}

Let

L=9⋅82k+1+5⋅62k−1.L=9\cdot8^{2k+1}+5\cdot6^{2k-1}.

Since k≥1k\geq1, both exponents are non-negative integers, so LL is an integer. Therefore,

f(k+1)−f(k)=7L.f(k+1)-f(k)=7L.

By the induction hypothesis,

f(k+1)= f(k)+7L= 7m+7L= 7(m+L).\begin{align*} f(k+1)=&\,f(k)+7L\\ =&\,7m+7L\\ =&\,7(m+L). \end{align*}

Hence f(k+1)f(k+1) is divisible by 7.

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, by mathematical induction, f(n)f(n) is divisible by 7 for every positive integer nn.