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FP1 2020 Oct Q8

A Level / Edexcel / FP1

IAL 2020 Oct Paper · Question 8

题目

Problem

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

 ∑r=1n2r2−1r2(r+1)2= n2(n+1)2\begin{align*} &\,\sum_{r=1}^{n}\frac{2r^2-1}{r^2(r+1)^2} \\ =&\,\frac{n^2}{(n+1)^2} \end{align*}
(6)

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

f(n)= 12n+ 2×5n−1\begin{align*} f(n)=&\,12^n \\ +&\,2\times5^{n-1} \end{align*}

is divisible by 77

(6)
题目中文翻译

8. (i) 用数学归纳法证明，对于 n∈Z+n \in \mathbb{Z}^+

 ∑r=1n2r2−1r2(r+1)2= n2(n+1)2\begin{align*} &\,\sum_{r=1}^{n}\frac{2r^2-1}{r^2(r+1)^2} \\ =&\,\frac{n^2}{(n+1)^2} \end{align*}
(6)

(ii) 用数学归纳法证明，对于 n∈Z+n \in \mathbb{Z}^+

f(n)= 12n+ 2×5n−1\begin{align*} f(n)=&\,12^n \\ +&\,2\times5^{n-1} \end{align*}

能被 77 整除。

(6)

解答

(i)

解法一

思路

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先验证 n=1n=1。归纳步骤中假设结论对 n=kn=k 成立，再给等式左边加上第 k+1k+1 项；通分后把分子展开成 (k+1)4(k+1)^4，即可得到 n=k+1n=k+1 时的目标式。

答题过程

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Let

Sn=∑r=1n2r2−1r2(r+1)2.S_n=\sum_{r=1}^{n}\frac{2r^2-1}{r^2(r+1)^2}.

For n=1n=1,

S1=2(1)2−112(1+1)2=14,S_1=\frac{2(1)^2-1}{1^2(1+1)^2}=\frac14,

and the right-hand side is

12(1+1)2=14.\frac{1^2}{(1+1)^2}=\frac14.

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

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

Sk=k2(k+1)2.S_k=\frac{k^2}{(k+1)^2}.

Adding the next term gives

Sk+1= k2(k+1)2+ 2(k+1)2−1(k+1)2(k+2)2.\begin{align*} S_{k+1} =&\,\frac{k^2}{(k+1)^2} \\ +&\,\frac{2(k+1)^2-1}{(k+1)^2(k+2)^2}. \end{align*}

The common denominator is

D=(k+1)2(k+2)2.D=(k+1)^2(k+2)^2.

The combined numerator is

N= k2(k+2)2+ 2(k+1)2−1= k4+4k3+4k2+ 2k2+4k+1= k4+4k3+6k2+ 4k+1= (k+1)4.\begin{align*} N=&\,k^2(k+2)^2 \\ +&\,2(k+1)^2-1 \\ =&\,k^4+4k^3+4k^2 \\ +&\,2k^2+4k+1 \\ =&\,k^4+4k^3+6k^2 \\ +&\,4k+1 \\ =&\,(k+1)^4. \end{align*}

Therefore,

Sk+1= ND= (k+1)4(k+1)2(k+2)2= (k+1)2(k+2)2.\begin{align*} S_{k+1} =&\,\frac ND \\ =&\,\frac{(k+1)^4}{(k+1)^2(k+2)^2} \\ =&\,\frac{(k+1)^2}{(k+2)^2}. \end{align*}

This is the required result with k+1k+1 in place of nn. 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, 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 f(1)f(1) 能被 77 整除。归纳步骤中把 f(k+1)f(k+1) 写成 12f(k)−14⋅5k−112f(k)-14\cdot5^{k-1}；归纳假设保证第一项能被 77 整除，第二项显然也含因子 77。

答题过程

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For n=1n=1,

f(1)=12+2×50=14,f(1)=12+2\times5^0=14,

which is divisible by 77.

Assume that the result is true for n=kn=k, where k∈Z+k\in\mathbb{Z}^+. Thus f(k)f(k) is divisible by 77.

Now

 f(k+1)= 12k+1+2×5k= 12(12k+2×5k−1)− 14×5k−1= 12f(k)− 14×5k−1.\begin{align*} &\,f(k+1) \\ =&\,12^{k+1}+2\times5^k \\ =&\,12\big(12^k+2\times5^{k-1}\big) \\ -&\,14\times5^{k-1} \\ =&\,12f(k) \\ -&\,14\times5^{k-1}. \end{align*}

By the inductive hypothesis, 12f(k)12f(k) is divisible by 77, and 14×5k−114\times5^{k-1} is also divisible by 77. Therefore, f(k+1)f(k+1) is divisible by 77.

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 77 for all n∈Z+n\in\mathbb{Z}^+ by mathematical induction.

The same recurrence can be derived by taking a difference. Let

Dk=f(k+1)−f(k).D_k=f(k+1)-f(k).

Then

Dk= 11×12k+8×5k−1= 11(12k+2×5k−1)− 14×5k−1= 11f(k)−14×5k−1.\begin{align*} D_k=&\,11\times12^k+8\times5^{k-1} \\ =&\,11\big(12^k+2\times5^{k-1}\big) \\ -&\,14\times5^{k-1} \\ =&\,11f(k)-14\times5^{k-1}. \end{align*}

Thus f(k+1)=f(k)+Dkf(k+1)=f(k)+D_k gives the same recurrence.

More generally, for any constant mm, set

Gm=f(k+1)−mf(k).G_m=f(k+1)-mf(k).

Expanding and regrouping,

Gm= (12−m)12k+ 2(5−m)5k−1= (12−m)12k+ 2(12−m)5k−1− 14×5k−1= (12−m)f(k)− 14×5k−1.\begin{align*} G_m=&\,(12-m)12^k \\ +&\,2(5-m)5^{k-1} \\ =&\,(12-m)12^k \\ +&\,2(12-m)5^{k-1} \\ -&\,14\times5^{k-1} \\ =&\,(12-m)f(k) \\ -&\,14\times5^{k-1}. \end{align*}

Taking m=1m=1 gives the difference above; taking m=12m=12 gives the recurrence directly. If the induction hypothesis is written as f(k)=7Mf(k)=7M for an integer MM, that recurrence becomes

 f(k+1)= 84M−14×5k−1= 7(12M−2×5k−1),\begin{align*} &\,f(k+1) \\ =&\,84M-14\times5^{k-1} \\ =&\,7\big(12M-2\times5^{k-1}\big), \end{align*}

which makes the divisibility by 77 explicit.

解法二

思路

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归纳框架不变，但在归纳步骤中计算 f(k+1)−5f(k)f(k+1)-5f(k)。含 5k5^k 的项恰好相消，剩下 7⋅12k7\cdot12^k，因而整除关系更直接。

答题过程

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For n=1n=1,

f(1)=12+2×50=14,f(1)=12+2\times5^0=14,

which is divisible by 77.

Assume that f(k)f(k) is divisible by 77 for some k∈Z+k\in\mathbb{Z}^+. Then

 f(k+1)−5f(k)= 12k+1+2×5k− 5(12k+2×5k−1)= 7×12k.\begin{align*} &\,f(k+1)-5f(k) \\ =&\,12^{k+1}+2\times5^k \\ -&\,5\big(12^k+2\times5^{k-1}\big) \\ =&\,7\times12^k. \end{align*}

Therefore,

f(k+1)= 5f(k)+ 7×12k.\begin{align*} f(k+1)=&\,5f(k) \\ +&\,7\times12^k. \end{align*}

Both terms on the right are divisible by 77, so f(k+1)f(k+1) is divisible by 77.

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 77 for all n∈Z+n\in\mathbb{Z}^+ by mathematical induction.