Treenewbee 发表于 2024-1-11 08:46 \[\sum_{n=1}^{300}{\frac{301-n}{4n^2-1}}=\frac{301}{2}*\sum_{n=1}^{300}{(\frac{1}{2n-1}-\frac{1}{2n+ ...
使用道具 举报
本版积分规则 发表回复 回帖并转播 回帖后跳转到最后一页
Archiver|手机版|小黑屋|数学中国 ( 京ICP备05040119号 )
GMT+8, 2025-6-20 15:39 , Processed in 0.079088 second(s), 14 queries .
Powered by Discuz! X3.4
Copyright © 2001-2020, Tencent Cloud.