这是陕西省第九次大学生高等数学竞赛复赛试题第九题

Question

这是陕西省第九次大学生高等数学竞赛复赛试题第九题能不能给个通俗易懂的方法

Answer 1

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