题目

Problem

Prove by induction that for all positive integers $n$

$\sum_{r=1}^{n} \log(2r - 1) = \log\left(\frac{(2n)!}{2^n n!}\right)$

(6)

题目中文翻译

用数学归纳法证明：对于所有正整数 $n$ ， $\sum_{r=1}^{n} \log(2r - 1) = \log\left(\frac{(2n)!}{2^n n!}\right)$

解答

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先验证 $n=1$ 。归纳步骤中，在前 $k$ 项的和后加上第 $k+1$ 项 $\log(2k+1)$ ，再用对数乘法法则合并。最后补出因子 $2k+2=2(k+1)$ ，将结果整理成题目在 $n=k+1$ 时的形式。

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For $n=1$ ,

\sum_{r=1}^{1}\log(2r-1)=\log1=0,

and

\log\left(\frac{2!}{2^1(1!)}\right) =\log1=0.

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

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

\sum_{r=1}^{k}\log(2r-1) =\log\left(\frac{(2k)!}{2^k k!}\right).

Then

\begin{align*} \sum_{r=1}^{k+1}\log(2r-1) =&\,\sum_{r=1}^{k}\log(2r-1) +\log(2k+1)\\ =&\,\log\left(\frac{(2k)!}{2^k k!}\right) +\log(2k+1)\\ =&\,\log\left( \frac{(2k)!(2k+1)}{2^k k!} \right)\\ =&\,\log\left( \frac{(2k+1)!}{2^k k!} \right). \end{align*}

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

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

Therefore,

\sum_{r=1}^{k+1}\log(2r-1) =\log\left( \frac{(2k+2)!}{2^{k+1}(k+1)!} \right),

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

\boxed{ \sum_{r=1}^{n}\log(2r-1) =\log\left(\frac{(2n)!}{2^n n!}\right) }

for all positive integers $n$ .

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官方评分资料也允许把归纳假设右侧拆成两个对数。加入第 $k+1$ 项后，先把分子合成 $(2k+1)!$ ；再把分子、分母同时补上 $2k+2$ ，即可分别得到目标中的新阶乘。

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The case $n=1$ is verified as in Method 1. Assume that the result is true for $n=k$ . Writing the logarithm as a difference,

\sum_{r=1}^{k}\log(2r-1) =\log(2k)!-\log(2^k k!).

Therefore,

\begin{align*} \sum_{r=1}^{k+1}\log(2r-1) =&\,\log(2k)!-\log(2^k k!)\\ &\,+\log(2k+1)\\ =&\,\log(2k+1)!-\log(2^k k!). \end{align*}

Now

(2k+2)!= (2k+2)(2k+1)!

and

2^{k+1}(k+1)! =(2k+2)2^k k!.

Hence

\begin{align*} &\,\log(2k+1)!-\log(2^k k!)\\ =&\,\log\left(\frac{(2k+2)!}{2k+2}\right)\\ &\,-\log\left( \frac{2^{k+1}(k+1)!}{2k+2} \right)\\ =&\,\log(2k+2)! -\log\big(2^{k+1}(k+1)!\big)\\ =&\,\log\left( \frac{(2k+2)!}{2^{k+1}(k+1)!} \right). \end{align*}

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

\boxed{ \sum_{r=1}^{n}\log(2r-1) =\log\left(\frac{(2n)!}{2^n n!}\right) }

for all positive integers $n$ .