这是陕西省第九次大学生高等数学竞赛复赛试题第九题
答案:1 悬赏:20 手机版
解决时间 2021-04-02 11:43
- 提问者网友：遮云壑
- 2021-04-01 12:25
这是陕西省第九次大学生高等数学竞赛复赛试题第九题
最佳答案
- 五星知识达人网友：煞尾
- 2021-04-01 12:43
2sin[π/(4n)]·sin[π/(4n)]
=1-cos[π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[π/(2n)]-cos[2π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[2π/(2n)]-cos[3π/(2n)]
……
2sin[π/(4n)]·sin[(2n-1)π/(4n)]
=cos[(n-1)π/(2n)]-cos[nπ/(2n)]
全部加起来，得到
2sin[π/(4n)]·∑sin[(2k-1)π/(4n)]
=1-cos[2π/(2n)]
=1
∴∑sin[(2k-1)π/(4n)]=1/{2sin[π/(4n)]}
令t=π/(4n)，则
原式=lim(t→0)π²/(16t²)·(1-t/sint)
=π²/16·lim(t→0)(sint-t)/(t²sint)
=π²/16·lim(t→0)(sint-t)/t³
=π²/16·lim(t→0)(cost-1)/(3t²)
=π²/16·(-1/6)
=-π²/96
=1-cos[π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[π/(2n)]-cos[2π/(2n)]
2sin[π/(4n)]·sin[3π/(4n)]
=cos[2π/(2n)]-cos[3π/(2n)]
……
2sin[π/(4n)]·sin[(2n-1)π/(4n)]
=cos[(n-1)π/(2n)]-cos[nπ/(2n)]
全部加起来，得到
2sin[π/(4n)]·∑sin[(2k-1)π/(4n)]
=1-cos[2π/(2n)]
=1
∴∑sin[(2k-1)π/(4n)]=1/{2sin[π/(4n)]}
令t=π/(4n)，则
原式=lim(t→0)π²/(16t²)·(1-t/sint)
=π²/16·lim(t→0)(sint-t)/(t²sint)
=π²/16·lim(t→0)(sint-t)/t³
=π²/16·lim(t→0)(cost-1)/(3t²)
=π²/16·(-1/6)
=-π²/96
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯