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FP1 2023 Jan Q9

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 9

题目

Problem

Prove by induction that for all positive integers nn

 ∑r=1nlog⁡(2r−1)= log⁡((2n)!2nn!)\begin{align*} &\,\sum_{r=1}^{n}\log(2r-1)\\ =&\,\log\Big(\frac{(2n)!}{2^n n!}\Big) \end{align*}
(6)
题目中文翻译

用数学归纳法证明：对于所有正整数 nn，

 ∑r=1nlog⁡(2r−1)= log⁡((2n)!2nn!)\begin{align*} &\,\sum_{r=1}^{n}\log(2r-1)\\ =&\,\log\Big(\frac{(2n)!}{2^n n!}\Big) \end{align*}
(6)

解答

解法一

思路

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先验证 n=1n=1。归纳步骤中，在前 kk 项的和后加上第 k+1k+1 项 log⁡(2k+1)\log(2k+1)，再用对数乘法法则合并。最后将分子、分母同乘 2k+2=2(k+1)2k+2=2(k+1)，把结果整理成题目在 n=k+1n=k+1 时的形式。

答题过程

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For n=1n=1,

 ∑r=11log⁡(2r−1)= log⁡1= 0,\begin{align*} &\,\sum_{r=1}^{1}\log(2r-1)\\ =&\,\log1\\ =&\,0, \end{align*}

and

 log⁡(2!21(1!))= log⁡1= 0.\begin{align*} &\,\log\Big(\frac{2!}{2^1(1!)}\Big)\\ =&\,\log1\\ =&\,0. \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

 ∑r=1klog⁡(2r−1)= log⁡((2k)!2kk!).\begin{align*} &\,\sum_{r=1}^{k}\log(2r-1)\\ =&\,\log\Big(\frac{(2k)!}{2^k k!}\Big). \end{align*}

Then

 ∑r=1k+1log⁡(2r−1)= ∑r=1klog⁡(2r−1) +log⁡(2k+1)= log⁡((2k)!2kk!) +log⁡(2k+1)= log⁡((2k)!(2k+1)2kk!)= log⁡((2k+1)!2kk!).\begin{align*} &\,\sum_{r=1}^{k+1}\log(2r-1)\\ =&\,\sum_{r=1}^{k}\log(2r-1)\\ &\,\hspace{2pt}+\log(2k+1)\\ =&\,\log\Big(\frac{(2k)!}{2^k k!}\Big)\\ &\,\hspace{2pt}+\log(2k+1)\\ =&\,\log\Big( \frac{(2k)!(2k+1)}{2^k k!} \Big)\\ =&\,\log\Big( \frac{(2k+1)!}{2^k k!} \Big). \end{align*}

Since 2k+2=2(k+1)2k+2=2(k+1),

 (2k+1)!2kk!= (2k+1)!2kk! ×2k+22k+2= (2k+2)!2kk!(2k+2).\begin{align*} &\,\frac{(2k+1)!}{2^k k!}\\ =&\,\frac{(2k+1)!}{2^k k!}\\ &\,\hspace{2pt}\times\frac{2k+2}{2k+2}\\ =&\,\frac{(2k+2)!}{2^k k!(2k+2)}. \end{align*}

The denominator can be written as

 2kk!(2k+2)= 2kk!⋅2(k+1)= 2k+1(k+1)!.\begin{align*} &\,2^k k!(2k+2)\\ =&\,2^k k!\cdot2(k+1)\\ =&\,2^{k+1}(k+1)!. \end{align*}

Therefore,

 ∑r=1k+1log⁡(2r−1)= log⁡((2k+2)!2k+1(k+1)!),\begin{align*} &\,\sum_{r=1}^{k+1}\log(2r-1)\\ =&\,\log\Big( \frac{(2k+2)!}{2^{k+1}(k+1)!} \Big), \end{align*}

which is the required result for n=k+1n=k+1. Hence, by mathematical induction,

 ∑r=1nlog⁡(2r−1)= log⁡((2n)!2nn!)\begin{align*} &\,\sum_{r=1}^{n}\log(2r-1)\\ =&\,\boxed{\log\Big(\frac{(2n)!}{2^n n!}\Big)} \end{align*}

for all positive integers nn.

解法二

思路

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也可以把归纳假设右侧拆成两个对数。加入第 k+1k+1 项后，先把分子合成 (2k+1)!(2k+1)!；再把分子、分母同时补上 2k+22k+2，即可分别得到目标中的新阶乘。

答题过程

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The case n=1n=1 is verified as in Method 1. Assume that the result is true for n=kn=k, where k∈Z+k\in\mathbb Z^+. Writing the logarithm as a difference,

 ∑r=1klog⁡(2r−1)= log⁡((2k)!) −log⁡(2kk!).\begin{align*} &\,\sum_{r=1}^{k}\log(2r-1)\\ =&\,\log\big((2k)!\big)\\ &\,\hspace{2pt}-\log(2^k k!). \end{align*}

Therefore,

 ∑r=1k+1log⁡(2r−1)= log⁡((2k)!) −log⁡(2kk!) +log⁡(2k+1)= log⁡((2k+1)!) −log⁡(2kk!).\begin{align*} &\,\sum_{r=1}^{k+1}\log(2r-1)\\ =&\,\log\big((2k)!\big)\\ &\,\hspace{2pt}-\log(2^k k!)\\ &\,\hspace{4pt}+\log(2k+1)\\ =&\,\log\big((2k+1)!\big)\\ &\,\hspace{2pt}-\log(2^k k!). \end{align*}

Now

 (2k+2)!= (2k+2)(2k+1)!\begin{align*} &\,(2k+2)!\\ =&\,(2k+2)(2k+1)! \end{align*}

and

 2k+1(k+1)!= 2k+1(k+1)k!= (2k+2)2kk!.\begin{align*} &\,2^{k+1}(k+1)!\\ =&\,2^{k+1}(k+1)k!\\ =&\,(2k+2)2^k k!. \end{align*}

Hence

 log⁡((2k+1)!) −log⁡(2kk!)= log⁡((2k+2)!2k+2) −log⁡(2k+1(k+1)!2k+2)= log⁡((2k+2)!) −log⁡(2k+2) −log⁡(2k+1(k+1)!) +log⁡(2k+2)= log⁡((2k+2)!) −log⁡(2k+1(k+1)!)= log⁡((2k+2)!2k+1(k+1)!).\begin{align*} &\,\log\big((2k+1)!\big)\\ &\,\hspace{2pt}-\log(2^k k!)\\ =&\,\log\Big(\frac{(2k+2)!}{2k+2}\Big)\\ &\,\hspace{2pt}-\log\Big( \frac{2^{k+1}(k+1)!}{2k+2} \Big)\\ =&\,\log\big((2k+2)!\big)\\ &\,\hspace{2pt}-\log(2k+2)\\ &\,\hspace{4pt}-\log\big(2^{k+1}(k+1)!\big)\\ &\,\hspace{6pt}+\log(2k+2)\\ =&\,\log\big((2k+2)!\big)\\ &\,\hspace{2pt}-\log\big(2^{k+1}(k+1)!\big)\\ =&\,\log\Big( \frac{(2k+2)!}{2^{k+1}(k+1)!} \Big). \end{align*}

Thus the result is true for n=k+1n=k+1. Since it is true for n=1n=1, by mathematical induction,

 ∑r=1nlog⁡(2r−1)= log⁡((2n)!2nn!)\begin{align*} &\,\sum_{r=1}^{n}\log(2r-1)\\ =&\,\boxed{\log\Big(\frac{(2n)!}{2^n n!}\Big)} \end{align*}

for all positive integers nn.