题目 Problem Prove by induction that for all positive integers nnn ∑r=1nlog(2r+1)=log((2n)!2nn!)\sum_{r=1}^{n} \log(2r + 1) = \log\left(\frac{(2n)!}{2^n n!}\right)∑r=1nlog(2r+1)=log(2nn!(2n)!) (6) 题目中文翻译 用数学归纳法证明：对于所有正整数 nnn， ∑r=1nlog(2r+1)=log((2n)!2nn!)\sum_{r=1}^{n} \log(2r + 1) = \log\left(\frac{(2n)!}{2^n n!}\right)∑r=1nlog(2r+1)=log(2nn!(2n)!) 解答
