✔ 最佳答案
圖片參考:
https://s.yimg.com/rk/AD06198854/o/3618015.png
2.作CP'⊥CP & CP'=CP則ΔACP≡ΔCP'B(SAS)獲得: CBP'=45度, BP'=AP=y又ΔCPQ≡ΔCQP'(SAS)可以得到: QP'=PQ=a在直角ΔQBP'裡面: QP'^2=QB^2+BP'^2=> PQ^2=BQ^2+AP^2
2014-09-25 17:19:21 補充:
1.
1/a1*a2=1/a1(a1+d)=[1/a1-1/(a1+d)]/d
1/a2*a3=1/a2(a2+d)=[1/a2-1/(a2+d)]/d
......
1/a99*a100=1/a99(a99+d)=[1/a99-1/(a99+d)]/d
Sum=s
={[1/a1-1/(a1+d)]+[1/a2-1/(a2+d)]+...+[1/a99-1/(a99+d)]}/d
={(1/a1-1/a2)+(1/a2-1/a3)+...+(1/a99-1/a100)}/d
=(1/a1-1/a100)/d
2014-09-25 17:21:27 補充:
=(a100-a1)/(d*a1*a100)
=(a1+99d-a1)/(d*a1*a100)
=99d/d*1*a100
=99/a100
=99/(1+99d)
1+99d=99/s
d(s)=(99-s)/99s
min=d(98)
=(99-98)/99*98
=1/9702